(i) The entire quantity of ammonia is consumed in the reaction. So ammonia is the Iimiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.
(ii) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide.
Amount of carbon dioxide leftover
= number of moles of CO₂ left over x molar mass of CO₂   
= 7 moles x 44 g mol^-1 
= 308 g

The balanced chemical equation is
\(\mathrm{MgCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_{2}\)
Molar mass of MgCO₃ is 84 g mol^–1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 g of MgCO₃ contain

= 28.57 g Mg.
i.e. percentage of magnesium
= 28.57 %.
84 g MgCO₃ contain 12 g of carbon

∴ 100 g MgCO₃ contain

= 14.29 g of carbon.
∴ Percentage of carbon
= 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 57.14 g of oxygen.
∴ Percentage of oxygen
= 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure MgCO₃ on heating gives 44 g of CO_2.
∴ 1000 g of 90 % pure MgCO₃ gives
\(\frac{\text { wt. of } \mathrm{MgCO}_{3}}{84 \mathrm{~g}} \frac{\text { % Purities }}{100 \%} \frac{\text { wt.of } \mathrm{CO}_{2}}{44 \mathrm{~g}}\)
\(x=44 \times \frac{100}{84} \times \frac{90}{100}\)
= 471.43 g CO₂
= 0.471 kg CO₂

Given:
Mg(s)+Br₂(I) ⟶ MgBr₂(s) ΔH_f⁰ = -524 KJ mol^-1
Sublimation:
Mg(s) ⟶ Mg(g), ΔH₁⁰ = +148 KJ mol^-1
Ionisation:
Mg(g) ⟶ Mg²⁺(g)+2e^- , ΔH₂⁰ =2187 KJ mol^-1
Vapourisation:
Br₂(I) ⟶ Br₂(g), ΔH₃⁰ = +31 KJ mol^-1
Dissociation:
Br₂(g) ⟶ 2Br(g), ΔH₄⁰ = +193 KJ mol^-1
Electron affinity:
Br(g) + e^- ⟶ Br^-(g), ΔH₅⁰ = -331 KJ mol^-1

ΔH_f = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + 2ΔH₅ + u
-524 = 148 + 2187 + 31 + 193 + (2 x -331) + u
-524 = 1897 + u
u = -524 - 1897
u = -2421 kJ mol^-1.

Electronic Configuration of Aluminium
\(\underbrace { { Al }^{ 13 }{ 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 1 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 13 - 9.5
(Z_eff)_Al =3.5
Electronic Configuration of chlorine
\(\underbrace { { 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 5 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 17 - 10.9
(Z_eff)_cl = 6.1
(Z_eff)_cl > (Z_eff)_Al and hence r_clAl

The required equation is
C₆H₁₂O_6(s)+6O_2(g) ➝ 6CO_2(g)+6H₂O_(l)
(i)  C_(graphite)+O_2(g) ➝ CO_2(g); ΔH= -395.0 KJ
(ii) H_2(g)+\(\frac{1}{2}\)O_2(g) ➝ H₂O_(l);  ΔH= -269.4 KJ
(iii) 6C_(graphite)+6H_2(g)+3O_2(g) ➝ C₆H₁₂O_6(s);  ΔH= -1169.8 KJ
Multiply equation (i) and (ii) by 6 and add them up
(iv) 6C_(graphite)+6H_2(g)+9O_2(g) ➝ 6CO_2(g)+6H₂O_(l) ; ΔH= -3984.6 KJ
Subtracting equation (iii) from (iv)
C₆H₁₂O_6(s)+6O_2(g) ➝ 6CO_2(g)+6H₂O_(l) ΔH= -2816.6 KJ
∴ Enthalpy of combustion of glucose = -2816.6 KJ.

CH₄+2O₂ ➝ CO2_(g)+2H₂O_(l)   ΔH^o= -890 KJ mol^-1
H₂O_(l) ➝ H₂O(g)     ΔH^o= 44 KJ mol^-1 at 298 K
ΔH^o for CH₄+2O_2(g) ➝ CO2_(g)+2H₂O_(g) 
=-890+44=-846 KJ mol^-1

 Ca_(s) + Cl_2(g)  \(\rightarrow \) CaCl2_(s)  ΔH_f^o = -795 KJ mol^-1
ΔH_f^o = ΔH₁^o + ΔH₂^o + ΔH₃^o + ΔH₄^o + ΔH₅^o 
ΔH₁^o for Ca_(s) \(\rightarrow \) Ca_(g)=121 KJmol^-1
ΔH₂^o for Ca_(g) \(\rightarrow \) Ca²⁺_(g) +2_e- = 2422 KJmol^-1
ΔH₃^o for Cl_2(g) \(\rightarrow \) 2Cl_(g) = +242.8 KJmol^-1
ΔH₄^o for 2Cl+2e- \(\rightarrow \)2Cl^- = 2 x -355 KJmol^-1 
ΔH_f^o = -795KJ mol^-1
ΔH_f^o = ΔH₁^o + ΔH₂^o+ ΔH₃^o + ΔH₄^o + ΔH₅^o 
ΔH_f^o = 121 + 2422 + 242.8[ 2 x 355] +ΔH₅^o 
-795 = 2785.8 - 710 + ΔH₅^o 
ΔH₅^o = 2870 KJ mol^-1 
\(\therefore\) Lattice enthalpy of CaCl₂ = 2870 KJ mol^-1

(i) T₁= 265 + 273 = 538 K
T₂ = 100 + 273 = 373 K
\(\eta =\left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right) \times 100\)
\(=\left( \frac { 538-373 }{ 538 } \right) \times 100\)
\(=\frac { 165 }{ 538 } \times 100\)
\(\eta =42.75%\)%
(ii) T₁ = 265 + 273 = 538 K
T₂ = 35 + 273 = 308 K
\(\eta =\left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right) \times 100\)
\( =\left( \frac { 538-308 }{ 538 } \right) \times 100\)
\(=\frac { 230 }{ 538 } \times 100\)
\(\eta =42.75%\)%.

\(\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{vap}}\)
=\(\frac{40850 J mole^{-1}}{373 K}\)
=109.51 JK^-1 mol^-1

Mass of He = 0.4 g; V = 10 Litres
Mass of O = 1.6 g; T = 27^oC + 273 = 300K
Mass of N = 1.4g
R = 0.0821 dm³ atm K^-1 mol^-1
Sol:
Calculate of no.of moles:
\(\left[ No.of\ moles=\frac { mass }{ molecular\ mass } \right] \)
No.of moles of He (n_He) = \(\frac { 0.4 }{ 4 } =0.1\)
No.of moles O₂ (\({ n }_{ { o }_{ 2 } }\)) = \(\frac { 1.6 }{ 32 } =0.05\)
No. of moles of N₂ (\({ n }_{ { N }_{ 2 } }\)) =\(\frac { 1.4 }{ 28 } =0.05\)
Total no.of moles (n) = n_He + \({ n }_{ { o }_{ 2 } }\) + \({ n }_{ { N }_{ 2 } }\)
= 0.1 + 0.05 + 0.05
n = 0.2
Calculate of mole fraction:
Mole fraction of He (X_He) = \(\frac { 0.1 }{ 0.2 } =0.5\)
Mole fraction of O₂  (\({ X }_{ { o }_{ 2 } }\)) =  \(\frac { 0.05 }{ 0.2 } =0.25\)
Mole fraction of N₂  ( \({ X }_{ { N }_{ 2 } }\)) = \(\frac { 0.05 }{ 0.2 } =0.25\)
Calculate of Total pressure of an ideal gas:
PV = nRT
\(P=\frac { nRT }{ V } \)
\(=\frac { 0.2\times 0.082\times 300 }{ 10 } =0.4926\)
P = 0.4926 atm
Calculation of Partial pressure:
Partial pressure of He (P_He) = 0.5 x 0.4926
= 0.2463 atm
[\(\therefore\) Partial pressure = mole fraction x Total pressure]
Partial pressure of O₂ (\({ P }_{ { o }_{ 2 } }\)) = 0.25 x 0.4926 = 0.1231 atm
Partial pressure of N₂ (\({ P }_{ { N }_{ 2 } }\)) = 0.25 x 0.4926 = 0.1231 atm
Calculation of Total pressure in the cylinder:
\(\therefore\) Total pressure P_Total = P_He + \({ P }_{ { o }_{ 2 } }\) + \({ P }_{ { N }_{ 2 } }\)
= 0.2463 + 0.1231 + 0.1231
= 0.4925
P_Total = 0.4925 atm

Given:
T_c = 33.2^oC + 273 = 306.2 K
P_C = 12.4 atm, R = 0.082 atm lit K^-1 mol^-1
Sol:
\({ T }_{ C }=\frac { 8a }{ 27Rb }\)...(1)
\({ P }_{ C }=\frac { a }{ { 27 }b^{ 2 } } \)...(2)
(1) + (2)
\(\frac { { T }_{ C } }{ { P }_{ C } } =\frac { 8a }{ 27Rb } \times \frac { { 27 }b^{ 2 } }{ a } =\frac { 8b }{ R } \)
\(\therefore \frac { 306.2 }{ 12.4 } =\frac { 8\times b }{ 0.082 } \)
\(b=\frac { 306.2\times 0.082 }{ 12.4\times 8 } =0.253\)
b = 0.253 lit mol^-1
Sub 'b' in (1)
\({ T }_{ C }=\frac { 8a }{ 27Rb } ;306.2=\frac { 8\times a }{ 27\times 0.082\times 0.253 } \)
\(a=\frac { 306.2\times 27\times 0.082\times 0.253 }{ 8 } =21.439\)
a = 21.439 atm lit² mol^-1

t₁ = 1.5 minutes (gas)_{hydro carbon}
t₂ = 4.73 minutes (gas)_Bromine
\(\frac { { \gamma }_{ Hydrocarbon } }{ { \gamma }_{ Bromine } } =\frac { { t }_{ Bromine } }{ { t }_{ Hydrocarbon } } \)
( \(\therefore\) Volume is constant)
\(=\frac { 4.73minutes }{ 1.5minutes } \)
=3.15
\(\frac { { \gamma }_{ Hydrocarbon } }{ { \gamma }_{ Bromine } } =\sqrt { \frac { { m }_{ Bromine } }{ { m }_{ Hydrocarbo } } } \)
\(3.15=\sqrt { \frac { 159.8{ gmol }^{ -1 } }{ { m }_{ Hydrocarbon } } } \)
= 3.15
Squaring on both sides and rearranging,
\({ m }_{ Hydrocarbon }=\frac { 159.8{ mol }^{ -1 } }{ { \left( 3.15 \right) }^{ 2 } } \)
m_hydrocarbon  = 16.1 g mol^-1
n(12) + (2n + 2) 1 = 16
12n + 2n + 2 = 16
14n = 16 - 2
14n = 14
n = 1
\(\therefore\) The hydrocarbon is C₁ H₂₍₁₎₊₂ = CH₄ 

Mass of He = 2.41g
No. of moles of He =\(\frac { Mass }{ MolarMass } \) =\(\frac { 2.41 }{ 4 } \)
= 0.6025 moles
Mass of Ne = 2.79 g
No. of moles of Ne = \(\frac { Mass }{ MolarMass } =\frac { 2.79 }{ 20 } \)
= 0.1395 moles
Volume of the Total no. of moles of the mixture
= 0.6025 + 0.1395 = 0.7420 moles
Container V = 1.04 dm³
Temperature T = 298 K
Pressure P = \(\frac { 1 }{ RT } \)RT
According to ideal gas equation PV = nRT
P=\(\frac { 0.7420x0.0821x298 }{ 1.04 } \)= 17.45 atm
Partial pressure P = mole fraction x Total pressure
=\(\frac { nA }{ nA+nB } \)  xp
Partial Pressure of Helium = PHc= \(\frac { 0.6025 }{ 0.7420 } x 17.45\)
According to Dalton's law of partial pressure
= 3.280 atm.
p = p1 + P2 + P3 
P_total = P_He + P_Ne = 14.169 + 3.280 = 17.449 atm

Mass of oxygen = 4.25 g
No. of moles of oxygen = \(\frac { 4.25 }{ 32 } =0.1328\) moles
Pressure P = 1.21 atm, volume V = 2.15 dm³
According to ideal gas equation
PV = nRT
\(T=\frac { PV }{ nR } =\frac { 1.21\times 2.15 }{ 0.1328\times 0.0821 } \)
\(\because\) R is the gas constant
R = 0.0821 L atm K mol^-1
\(=\frac { 2.6015 }{ 0.0109 } =238\quad K\)

(a) 5.076 atm
(b) 3.495 atm

Volume of the neon gas V₁ = 375 ml
of a temperature T₁ = 78°C + 273
= 351 K
At a temperature of T₂ = 22°C + 273
= 295 K
The volume of neon gas V₂ = ? 
According to Charles law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } } \)
\({ V }_{ 2 }=\frac { 375\times 295 }{ 351 } =315.17\quad ml\)
\(\therefore\)  Volume of neon gas at a temperature of 22°C = 315.17 ml 

Given:
R=8.314 JK^-1mol^-1
T=278K
K=1.8 x 10^-7
ΔG⁰=-2.303 RT log K
ΔG⁰=-2.303 x 8.314 x log(1.8 x 10^-7)
ΔG⁰=-2.303 x 8.314 x 298 x (-6.745)
ΔG⁰=38484 J mol^-1
ΔG⁰=38.484 KJ mol^-1.

Given:
ΔH⁰=-990 kJ mol^-1=-990000 J mol^-1
ΔS⁰=-342.98 JK^-1; T=25⁰C=298K
ΔG⁰=ΔH⁰-TΔS⁰
=-990000-298(-342.98)
=-990000+102208.04
=-887791.96 mol^-1
ΔG⁰=-887.791 kJ mol^-1

Given: ΔH⁰ _reaction=179.9 KJ mol-1=179900 J mol-1, ΔS⁰_reaction=78.09 JK-1 mol-1, Temperature, T=300K, T=25⁰C=298 K
ΔG⁰=ΔH⁰-TΔS⁰
=179900-300(78.09)J mol^-1
=179900-23,427 J mol^-1
=156473 J mol^-1
 ΔS⁰_reaction=156.473 KJ mol^-1

\({ \triangle S }_{ vap }=\frac { { \triangle H }_{ vap } }{ { T }_{ b }(K) } =\frac { 40850J/mol }{ 373K } \)=109.517 JK^-1mol^-1

Given: ΔH_f⁰, CO₂=-96200 cal; ΔH_f⁰, H₂O=-68000 cal, ΔH_C⁰, C₂H₅OH=34,600 cal
C₂H₅OH_(I) + 3O_2(g) ⟶ 2CO_2(g) + 3H₂O_(I)
\({ \triangle H }^{ 0 }={ \sum { H } }_{ Products }^{ 0 }-{ \sum { H } }_{ Reactants }^{ 0 } \)
=(-2 x 96200) + (-3 x 68000)-34600
=-192400-204000 -34600=-431000
ΔH⁰=-431000 cal

w=250 J
q =100 J
[Work done on the system, w > 0 Heat given out of the system, q < 0]
ΔU=q+w
=(-100)+(+250)=50
ΔU=+50J

2KCIO_(s) \(\longrightarrow\) 2KCI_(s) + 3O_2(g)
P_total = 772 mm Hg
\({ P }_{ { H }_{ 2 }O }\) = 26.7 mm Hg
P_total = \({ P }_{ { O }_{ 2 } }+{ P }_{ { H }_{ 2 }O }\)
\(\therefore { P }_{ { O }_{ 2 } }={ P }_{ total }-{ P }_{ { H }_{ 2 }O }\)
P₁ = 26.7 mm Hg, T₂ = 295 K
T₁ = 300k, P₂ = ?
\(\frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow { P }_{ 2 }=\left( \frac { { P }_{ 1 } }{ { T }_{ 1 } } \right) { T }_{ 2 }=\frac { 26.7mm\ Hg }{ 300K } \times 295K\)
P₂ = 26.26 mm Hg
\(\therefore { P }_{ { O }_{ 2 } }\)= 772 - 26.26 = 745.74 mm Hg.

Vo₂ = 12 dm³
V_He = 46 dm³
V_total = 5 dm³
\(\left.\begin{array}{c} T=298 \mathrm{~K} \\ \mathrm{P}=1 \mathrm{~atm} \end{array}\right\} \text { Constant }\)
P_O2 = X₀₂ x P_total 
\( \mathrm{X}_{\mathrm{O}_{2}}=\frac{\mathrm{n}_{\mathrm{O}_{2}}}{\mathrm{n}_{\mathrm{O}_{2}}+\mathrm{n}_{\mathrm{He}}} \)
\(\mathrm{n}_{\mathrm{o} 2}=0.54 \mathrm{~mol} \)
\(=\frac{0.54}{0.54+2.05} \)
\(=\frac{0.54}{2.59}=0.21\)

n_He = 2.05mol
P_total x V_total = 1 atm x 22.4 l
\(\therefore { P }_{ total }=\frac { 1atm\times 22.4l }{ 5l } \)
P_total = 4.48 atm
= 0.94 atm
P_He = X_He x P_total
\({ X }_{ He }=\frac { { n }_{ He } }{ { n }_{ o2 }+{ n }_{ He } } =\frac { 2.05 }{ 0.54+2.05 } \)
\({ X }_{ He }=\frac { 2.05 }{ 2.59 } =0.79\)
\(\therefore\) P_He  = 0.79 x 4.48 atm
P_He = 3.54 atm.

For 1 mole of an ideal gas,
PV=RT
P=4.1 atm
V=2lt.
PV=RT
T=\(\frac{PV}{R}=\frac{4.1atm\times2lit\times1mole}{0.082 lit atm K^{-1}mol^{-1}}\)
=100 K
\(\Delta S=\frac{q}{T}\)
\(\Delta S=\frac{3710J}{100K}=37.1 JK^{-1}\)
ΔS of expansion = 37.1 JK^-1.

T_system=273+55=328 K
T_surroundings=20+273=293 K
ΔS_Univ=ΔS_total=ΔS_system+ΔS_surroundings
\(\Delta S_{system}=\frac{q_{system}}{T_{system}}=\frac{-75J}{328K}=-0.2287 JK^{-1}\)
\(\Delta S_{surroundings}=\frac{q_{Surroundings}}{T_{Surroundings}}=\frac{+75J}{293K}=0.260 JK^{-1}\)
ΔS_Univ=ΔS_total=[-0.2287+0.26]
=0.0313 JK^-1.

% Efficiency = \([\frac{T_1-T_2}{T_1}]\times100\)
T₁=110+273=383 K.
T₂=25+273=298 K.
% Efficiency = \([\frac{383-298}{383}]\times100\)
=22%.

Given: \({ S }_{ { I }_{ 2 } }^{ o }=118.7\quad J{ K }^{ -1 }{ mol }^{ -1 };{ S }_{ { HI } }^{ o }=206.3\quad J{ K }^{ -1 }{ mol }^{ -1 };{ S }_{ { { H }_{ 2 } } }^{ o }=140.6\quad J{ K }^{ -1 }{ mol }^{ -1 }\)
Formula : \({ \Delta S }^{ O }={ S }_{ HI }^{ o }-\frac { 1 }{ 2 } ({ S }_{ { H }_{ 2 } }^{ o }+{ S }_{ { I }_{ 2 } }^{ o })\)
ΔG^o=ΔH^o-TΔS^o
\(\Delta H_{f}^{o}\)=+25.955 kJ mol^-1
ΔS^o=\(\Sigma S_{Products}^{o}-\Sigma _{Reactants}^{o}\)
=206.3-\(\frac{1}{2}\)(140.6+118.7)
=206.3-129.65
ΔS^o=76.65 kJ^-1mol^-1
ΔG=ΔH-TΔS
=25.95-(373ｘ76.65ｘ10^-3)
=25.95-28.59
=-2.640 kJ^-1 mol^-1
When ΔG is -ve.The reaction is spontaneous.

\({ \Delta G }_{ reaction}^{ o }={\Sigma \Delta G }_{ products }^{ o }-{ \Sigma\Delta G }_{ reactants }^{ o }\)
=2(1.8)-(-33.8)
=3704 kJ mol^-1
\({ \Delta G }_{ reaction }^{ o }\)=+37.4 kJ mol^-1
Since \({ \Delta G }^{ o }\) is positive, the reaction will not proceed spontaneously in the forward direction.

Volume of the gas in the balloon V₁ = 400 ml
temperature T₁ = -120 °C + 273
= 153 K
If the balloon is warmed T₂  = 100° C + 273
to a temperature = 373 K
Then the volume of the gas V₂ = ?
According to charles law \(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\frac { { V }_{ 2 } }{ { T }_{ 1 } } \times { T }_{ 2 }\)
\(=\frac { 400 }{ 153 } \times 373=975\quad ml\)
\(\therefore\)   Volume of helium gas in the balloon at a temperature of 100°C = 975 ml

According to charles law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { 1.54 }{ 293 } \)
= 0.0052
T₁ = 20 C + 273 = 293 K
T₂ = 40°C
\(\frac { { V }_{ 2 } }{ { T }_{ 2 } } =\frac { 1.65 }{ 40+273 } =\frac { 1.65 }{ 313 } =0.0052\)
T₃ = 100°C + 273 = 373 K
\(\frac { { V }_{ 3 } }{ { T }_{ 3 } } =\frac { 1.95 }{ 373K } =0.0052\)
T₄ = 120°C+273 = 393 K
\(\frac { { V }_{ 4 } }{ { T }_{ 4 } } =\frac { 2.07 }{ 393 } =0.0052\)
The average value of constant of proportionality is 0.0052.

Initial temperature T₁ of the gas = 0°C = 0 + 273
= 273°C
Let the Initial volume of the gas V₁ = x ml
The final volume V₂ of the gas = 1.25 \(\times\) xml
According to Charles' law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\(\frac { x }{ 273 } =\frac { 1.25\times x }{ { T }_{ 2 } } \)

= 341.25 K
The final temperature = 341.25 K 

At sea level pressure P₁ = 1 atm
Volume occupied at sea level V₁ = 785 \(\times\) 10^-3 dm³
If the pressure P₂ = 0.052 atm
the volume of the balloon V₂  = ?
According to Boyle's law
P₁ V₁ = P₂ V₂
1 \(\times\) 785 \(\times\) 10^-3 = 0.052 \(\times\) V₂
\({ V }_{ 2 }=\frac { 1\times 785\times { 10 }^{ -3 } }{ 0.052 } \)
= 15096.15 \(\times\) 10^-3 dm³

Volume of gas V ₁ = 6.85 dm³
at a pressure P₁ = 0.650 atm
If the pressure increased
P₂ by 0.5 atm = P₁ + 0.5
= 0.650 + 0.5
= 1.15 atm
Volume of the gas V₂ = ?
According to Boyle's law
P₁ V₁ = P₂ V₂,
6.85 \(\times\) 0.650 = V₂  \(\times\) 1.15
V₂ = \(\frac { 6.85\times 0.650 }{ 1.15 } \)
\(=\frac { 4.4525 }{ 1.15 } \)
= 3.871 dm³

Volume of oxygen V₁ = 2.5 L
at a pressure P₁ = 5.0 atm
At pressure P₂ = 1.01 atm
Volume of oxygen V₂ = ?
According to Boyle's law
P₁ V₁ = P₂ V₂
5 \(\times\) 25 = V₂ \(\times\) 1.01
\({ V }_{ 2 }=\frac { 12.5 }{ 1.01 } =12.37\quad L\)
Volume of oxygen at a pressure of 1.01 atm
= 12.37 l

ΔG⁰ = 2.303 RT log K_p
where
R = 8.314 JK^-1 mol^-1;
K_p = 2.47 x 10^-29;
T=298K
ΔG⁰ = -2.303 (8.314 JK^-1 mol^-1) (298K)log(2.47 x 10^-29)
ΔG⁰ = 16300 J mol^-1
ΔG⁰ = 16.3 kJ mol^-1

\(CO+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { CO }_{ 2 }\)
\({ \triangle }G_{ (reaction) }^{ 0 }={ \sum { G } }_{ f(products) }^{ 0 }-{ \sum { G } }_{ f(reactants) }^{ 0 }\)
\({ \triangle G }_{ (reaction) }^{ 0 }=\left[ { G }_{ { CO }_{ 2 } }^{ 0 } \right] -\left[ { G }_{ CO }^{ 0 }+\frac { 1 }{ 2 } { G }_{ { O }_{ 2 } }^{ 0 } \right] \)
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-394.4+[137.2+0]
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-257.2 kJ mol^-1
\({ \triangle G }_{ (reaction) }^{ 0 }\) of a reaction at a given temperature is negative hence the reaction is spontaneous.

ΔH_t = 2090 Jmol^-1
T_t = 13+273 = 286K
ΔS_t = \(\frac { { \triangle H }_{ t } }{ { T }_{ t } } \)
ΔS_t = \(\frac { 2090 }{ 286 } \)
ΔS_t = 7.307 JK^-1mol^-1​​​​​​​

Pressure of a gas at ground level P₁ = 0.965 atm
Volume of the gas at ground level V₁ = 10.0 m³
Volume of the gas at an altitude of 5300m V₂  = 20.0 m³
Pressure of the gas at an altitude of 5300 P₂ = ?
According to Boyle's law = T₁ V₁ = P₂ V₂
0.0965 \(\times\) 10 = P₂ \(\times\) 20.0
\({ P }_{ 2 }=\frac { 0.0965\times 10.0 }{ 20.0 } \)
= 0.04825 atm
Pressure of the gas at an altitude of 5300m
P₂  = 0.04825 atm

According to Graham's law of diffusion
\(\frac { { r }_{ H_{ 2 } } }{ { r }_{ K_{ r } } } =\sqrt { \frac { { M }_{ Kr } }{ { M }_{ { H }_{ 2 } } } } \)
\(\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { K }_{ r } } } =\sqrt { 42 } \)
\(\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ Kr } } =6.480\)
\({ r }_{ { H }_{ 2 } }=6.480\quad { r }_{ Kr }\)
1 : 6480 is ratio of H₂ : Kr

Molar mass of N₂ = 28 g/ mole
rate of diffusion = 192 sec
of unknown gas rate of diffusion of \({ r }_{ N_{ 2 } }\) (nitrogen gas) = 84 sec
According to Graham's law or
\(\frac { { r }_{ unknown } }{ { r }_{ { N }_{ 2 } } } =\sqrt { \frac { { M }_{ { N }_{ 2 } } }{ { M }_{ unknown } } } \)
Molecular mass of unknown gas
\({ M }_{ unknown }=\frac { { M }_{ N_{ 2 } }\times { \left( { V }_{ { N }_{ 2 } } \right) }^{ 2 } }{ { \left( { { r }_{ unknown } } \right) }^{ 2 } } \)
\({ M }_{ unknown }=\frac { 28\times 84\times 84 }{ 192\times 192 } \)
\(=\frac { 197568 }{ 36864 } =5.35\quad g/mole\)

0.15 dm²

Given data V₁ = 275 ml n₁ = 0.120 moles
V₂ = ?
\({ n }_{ 1 }=\frac { mass }{ molecular\ mass } =\frac { 1 }{ 44 } \)
Assuming pressure and temperature are constant According to Avogadro's hypothesis,
n₂ = 0.227 moles
\(\frac { { V }_{ 1 } }{ { n }_{ 1 } } =\frac { { V }_{ 2 } }{ { n }_{ 2 } } \)
\({ V }_{ 2 }=\frac { { V }_{ 1 }{ n }_{ 2 } }{ { n }_{ 1 } } \)
\({ V }_{ 2 }=\frac { 275\times 0.0227 }{ 0.120 } \)
V₂ = 52 moles
The volume of the balloon after dry ice has vaporized is V₁ + V₂ = 275 + 52 = 327 ml.

ΔG = ΔH - TΔS
T = \(\frac{\Delta H-\Delta G}{\Delta S}\)=\(\frac{30560-0}{66}\)=463
T = 463K.
At 463 K, the reaction is at equilibrium
At T < 463 K, ΔG will have positive value hence backward reaction is favoured.

= -290830 - 298 (-330)
= -290830 + 98340 = -192490
\(\Delta{G^{o}}\)=-192490 J mol^-1
Since \(\Delta{G^{o}}\) is negative, the reaction is spontaneous. 

C+O₂ ➝ CO₂     \(\Delta { S }_{ f }^{ o }\)=?
\(\Delta { S }_{ f }^{ o }\), CO₂ = \({ \Sigma S }_{ Compound }^{ o }-{ \Sigma S }_{ elements }^{ o }\)
=218.8(8.74+205.60)
\(\Delta { S }_{ f }^{ o }\), CO₂ =4.46 Jk^-1

\(\Delta n\)=10-12=-2 mol

=-4984ｘ10³J+8.314 JK^-1 mol^-1ｘ298kｘ(-2)mol.
=-4984000 J-4955.144J=-4988955.144J
\(\Delta H\)=-4988.955 kJ

\(\Delta{S}_{univ}=\Delta{S}_{sys}+\Delta{S}_{surr}\)
=-22.88+(+24.85)
=1.97 Jk^-1mol^-1
\(\therefore \underline { \Delta } { S }_{ Univ }>0\) at 272K
∴ The process of freezing of water is spontaneous

ΔH_fusion = 6008 J mol-1
Tf=0⁰C = 273K
\({ H }_{ 2 }O_{ (S) }\quad \overset { 273K }{ \longrightarrow } \quad { H }_{ 2 }O_{ (I) }\)
ΔS_(fusion) = \(\frac { \triangle H_{ fusion } }{ { T }_{ f } } \)
ΔS_(fusion) = \(\frac { 6008 }{ 273 } \)
ΔS_(fusion) = 22.007 JK^-1 mole^-1

C(g) + O2(g) ⟶ CO2(g)
ΔS_r⁰ = \({ \sum { S } }_{ (Products) }^{ 0 }-{ \sum { S } }_{ (reactants) }^{ 0 }\)
ΔS_r⁰ = \(\left\{ { S }_{ CO2 }^{ 0 } \right\} -\left\{ { S }_{ C }^{ 0 }+{ S }_{ O2 }^{ 0 } \right\} \)
ΔS_r⁰ = 213.6 - [5.74+205]
ΔS_r⁰ = 213.6 - [210.74]
ΔS_r⁰ = 2.86 JK^-1

q = 957.7 kJ
T = 110+273 = 383 K
ΔS= \(\frac { { q }_{ rev } }{ T } \)
ΔS = \(\frac { 957.7 }{ 383 } \)
ΔS =2.5 kJ K^-1

% Efficiency =\(\left[ \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right] \times 100\)
Here
T₁ = 816+273=1089 K;
T₂ = 21+273 = 294 K
% Efficiency= \(\left( \frac { 1089-294 }{ 1089 } \right) \times 100\)
% Efficiency = 73%

ΔH_c(CH₄)= - 890 kJ mol^-1
ΔH_c(C₂H₄) = -1423 kJ mol^-1
ΔH_c=-203.87 kJ mol^-1
Let the mixture contain x lit of CH₄ and (3.67 - x)
lit of ethylene
CH₄+2O₂\(\rightarrow \)2CO₂+2H₂O
XLit
C₂H₄+3O₂\(\rightarrow \)2CO₂+2H₂O
(3.67-X)Lit       2(3.67-X)Lit
Volume of Carbondioxide formed
=x + 2 (3.67 - x) = 6.11 lit
X+7.34-2X=6.11
7.34-X=6.11
X=1.23Lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence
\(\triangle { H }_{ C }=\left[ \frac { \triangle { H }_{ C }\left( { CH }_{ 4 } \right) }{ 22.4Lit } \times \left( X \right) lit \right] +\left[ \frac { \triangle { H }_{ C }\left( { C }_{ 2 }{ H }_{ 4 } \right) }{ 22.4lit } \times \left( 3.67-X \right) lit \right] \)
\(\triangle { H }_{ C }=\left[ \frac { -890KJ{ mol }^{ -1 } }{ 22.4Lit } \times 1.23lit \right] +\left[ \frac { -1423 }{ 22.4lit } \times \left( 3.67-1.23 \right) lit \right] \)
ΔH_c =[-48.87kJ mol^-1]+[-155kJ mol^-1]
ΔH_c =-203.87kJ mol^-1 .

T=175⁰ C=175+273=448K
Concentration of l-pentyne [A] = 1.3%
Concentration of2-pentyne [B] = 95.2%
Concentration of 1, 2-pentadiene [C] = 3.5%
At equilibrium
B\(\rightleftharpoons \)A
95.2%1.3%\(\Rightarrow \)
\({ K }_{ 1 }=\frac { 1.3 }{ 95.2 } =0.0136\)
B\(\rightleftharpoons \)​​C
95.2%3.5%\(\Rightarrow \)
\({ K }_{ 2}=\frac { 3.5 }{ 95.2 } =0.0367\)
\(\Rightarrow \)\(\Delta { G }_{ 1 }^{ 0 }\) =-2.303RT log K₁
\(\Delta { G }_{ 1 }^{ 0 }\)=-2.303X8.314X448Xlog0.0136
\(\Delta { G }_{ 1 }^{ 0 }\) =+16010J
\(\Delta { G }_{ 1 }^{ 0 }\)=+16 kJ
\(\Rightarrow \)\(\Delta { G }_{ 2 }^{ 0 }\)=-2.303X8.314X448Xlog0.0367
\(\Delta { G }_{ 2 }^{ 0 }\) =+12312J
\(\Delta { G }_{ 2 }^{ 0 }\)=+12.312 kJ.

ΔH_f(Fe₂O₃)= -741 kJ mol^-1
ΔH_f(CO)= -137 kJ mol^-1
ΔH_f(CO₂)= -394.5 kJ mol^-1
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂ ΔH_r=?
ΔH_r=Σ(ΔH_f)products - Σ(ΔH_f)reactants
ΔH_r=[2ΔH_f=(Fe)+3ΔH_f(CO₂)]-[ΔH_f(Fe₂O₃)+3ΔH_f(CO)]
ΔH_r=[0 + 3 (-394.5)] - [-741 +3 (-137)]
ΔH_r=[-1183.5] - [-1152]
ΔH_r=-1183.5 + 1152
ΔH_r=-31.5 kJ mol^-1

Given data P₁ = 1.05 atm V₁ = 0.475 dm³
P₂ = 5.65 atm V₂ = ?
P₁V₁ = P₂V₂
\({ V }_{ 2 }=\frac { { P }_{ 1 }V_{ 1 } }{ { P }_{ 2 } } \)
\({ V }_{ 2 }=\frac { 1.05\ atm\times 0.475{ dm }^{ 3 } }{ 5.65\ atm } \)
V₂ = 0.088 dm³

Given data P₁ = 1 atm V₁ = 5.82 dm³
P₁ = 1.92 atm V₂ = ?
According to Boyles law,
P₁V₁ = P₂V₂
\({ V }_{ 2 }=\frac { { P }_{ 1 }{ V }_{ 1 } }{ { P }_{ 2 } } \)
\({ V }_{ 2 }=\frac { 1\ atm\times 5.82\ { dm }^{ 3 } }{ 1.92\ atm } \)
V₂ = 3.031 dm²

a = 2.095 lit² atm mol^-1; R = 0.0821 dm³ atm lit K^-1 mol^-1
b = 0.0189 lit mol^-1
\({ T }_{ i }=\frac { 2a }{ Rb } \)
\(=\frac { 2\times 2.095 }{ 0.0821\times 0.0189 } =2700.28\quad K\)
T_i = 2700.28 K

a = 3.67 atm lit² mol^-2
b = 0.0408 lit mol^-1
R = 0.0821 atm lit K^-1 mol^-1
(i) \({ T }_{ c }=\frac { 8a }{ 27Rb } \)
\(=\frac { 8\times 3.67 }{ 27\times 0.082\times 0.0408 } =324.7\quad K\)
T_c = 324.7 K
(ii) \({ P }_{ c }=\frac { a }{ 27{ b }^{ 2 } } \)
\(=\frac { 3.67 }{ 27\times { (0.0408) }^{ 2 } } =81.6\)
P_c = 81.6 atm.

a = 6.34 atm lit^-2; b = 52.6 ml mol^-1
b = 0.056 lit mol^-1
\({ T }_{ c }=\frac { 8a }{ 27Rb } \)
\(=\frac { 8\times 6.34 }{ 27\times 0.0821\times 0.0526 } =434.997\simeq 435\quad K\)
T_C = 162 °​​​​​​C
\({ P }_{ C }=\frac { a }{ { 27b }^{ 2 } } =\frac { 6.34 }{ 27\times { (0.0526) }^{ 2 } } =84.87\)
P_c=\(\frac { a }{ 27{ b }^{ 2 } } =\frac { 6.34 }{ 27\times { (0.0526) }^{ 2 } } =84.87\)
P_c = 84.87 atm.

Time taken for effusion of gas A = 73 seconds
Time taken for effusion of SO₂ gas = 75 seconds
Volume of gases (A & SO₂) taken = 75 ml each.
Molecular mass of SO₂ =
\(\frac { Effusion\ rate\ of\ { SO }_{ 2 } }{ Effusion\ rate\ of\ A } =\sqrt { \frac { { M }_{ A } }{ { M }_{ { SO }_{ 2 } } } } \)
\(\frac { 75/75 }{ 75/73 } =\sqrt { \frac { { M }_{ A } }{ { M }_{ { SO }_{ 2 } } } } ;\frac { 1 }{ 1.027 } =\sqrt { \frac { { M }_{ A } }{ 64 } } \)
\({ (0.9737) }^{ 2 }=\frac { { M }_{ A } }{ 64 } ;{ M }_{ A }=60.67\)
\(\therefore\)  Molecular mass of A is 60.67.

The first element of the periodic table is hydrogen.
2H₂ + O₂ ➝ 2H₂O
The compound formed is water.
Auto - ionization or auto-protolysis of water proceeds as follows
\(\underset { Acid }{ { H }_{ 2 }O(l) } +\underset { Base }{ { H }_{ 2 }O(l) } \longleftrightarrow \underset { Conjugate\quad Acid }{ { H }_{ 3 }O } +(aq)+\underset { Conjugate\quad Acid }{ OH-(aq) } \)

2H₂ + O₂ ➝ 2H₂O
The isotope of hydrogen which has one proton and one neutron in its nucleus is Deuterium.
2D₂ + O₂ ➝ 2D₂O
The product is heavy water (Deuterium oxide). H₂O and D₂O have same chemical properties but the
reaction velocity of D2O is slightly less due to the difference in the mass number of the isotopes known as isotopic effect. Deuterium is heavier than protium so reacts slowly.

(i) The element (A) is hydrogen reacts with carbon monoxide to form methanol (B).
CO + 2H₂ \(\overset{Cu}{\rightarrow}\) CH₃OH
         (A)           (B)
(ii) Hydrogen reacts with nitrogen to form ammonia (C). 
\({ N }_{ 2 }+3{ H }_{ 2 }\quad \overset { 380-{ 450 }^{ 0 }C }{ \underset { 200/atm\quad 1le }{ \rightleftharpoons } } 2\underset { (C) }{ { NH }_{ 2 } } \)

(i) A important peroxide is hydrogen peroxide (A).
(ii) H₂O₂ disproportionates to form oxygen and water.
\(\underset { (A) }{ { H }_{ 2 }{ O }_{ 2 } } \quad \longrightarrow { H }_{ 2 }O+\frac { 1 }{ 2 } { O }_{ 2 }\)
(iii) H₂O₂ reacts with FeSO₄ in acidic condition to form ferric sulphate (B).
2FeSO₄ + H₂SO₄ + H₂O₂ ➝ Fe₂(SO₄)₃ + 2H₂O
                                (A)           (B)
(iv) H₂O₂ reacts with KMnO₄ in basic condition and form Manganese dioxide (C) and potassium hydroxide (D).
2KMnO₄ + 3H₂O₂  ➝ 2MnO₂ + 2KOH + 2H₂O + 3O₂
                   (A)            (C)          (D)

(i) The universal solvent is water (A).
(ii) Water reacts with chlorine gas to form Hydrochloric acid (B) and Hypochlorous acid (C).
Cl₂ + H₂O ➝ HCl + HOCl
          (A)       (B)     (C)
(iii) Water reacts with silicon tetrachloride (SiCI₄) to give silicon dioxide (D).
SiCl₄ + 2H₂O  ➝ SiO₂ + 4HCl
               (A)          (D)     (B)

(i) An element (A) belonging to group number 1 and period number 3 is sodium (A).
(ii) Sodium reacts with hydrogen to form sodium hydride (B)
2Na + H2 ➝ 2NaH
   (A)  (B)       (C)
(iii) Sodium reacts with water to form sodium hydroxide (C).
2Na + 2H₂O ➝ 2NaOH + H₂
  (A)                    (C)

Applying Graham's law of Diffusion,
\(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } ;\frac { 1/2 }{ 1 } =\sqrt { \frac { 32 }{ { M }_{ 1 } } } \)
Squaring both sides of the equation,
\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 32 }{ { M }_{ 1 } } ;\frac { 1 }{ 4 } =\frac { 32 }{ { M }_{ 1 } } \)
\(\therefore\) M1 = 128
Thus the molecular mass of the unknown gas is 128.

Mass of neon = 167.5
Mass of dioxygen = 60.8 g
Pressure = 20 bar
No. of moles of dioxygen = \(\frac { 60.8\ g }{ 32\ g\ mol } =1.9\ mol\)
No. of moles of neon = \(\frac { 167.5\ g }{ 20\ g\ mol } =8.375\ mol\)
Mole fraction of dioxygen  \(=\frac { 1.9 }{ 1.9+8.375 } =\frac { 1.9 }{ 10.275 } =0.185\)
Mole fraction of neon = \(\frac { 8.375 }{ 1.9+8.375 } =\frac { 8.375 }{ 10.275 } =0.815\)
Partial pressure of gas = Mole fraction \(\times\)  x Total pressure
Partial pressure of dioxygen = 0.185 \(\times\) 20 = 3.7 bar
Partial pressure of neon = 0.815 \(\times\) 20 = 16.3 bar.

P₁ = 4 bar; P₂ = ?
V₁ = 2 litre; V₂ = 4 litre
T₁= 27 + 273 = 300 K
T₂ = 77 + 273 = 350 K
\(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \)
\({ P }_{ 2 }=\frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \times \frac { T_{ 2 } }{ V_{ 2 } } =\frac { 4\times 2\times 350 }{ 300\times 4 } =2.33\ bar\)
The pressure of CO gas will be 2.33 bar.

Density (d) = 0.9 gm lit^-1
T = 350 K
Mass of neon (M) = 20 gm mol^-1
\(P=\frac { dRT }{ M } \quad \left( \because n=\frac { d }{ M } \right) \)
\(=\frac { 0.9\times 8.314\times 10^{ -2 }\times 350 }{ 20 } =1.309\quad bar\)
Pressure of neon = 1.309 bar

P = 250 bar; T = 350 K; V = 500 ml = 0.5 litre
\(P=\frac { nRT }{ V } [\because PV=nRT]\)
\(=\frac { 250\times 0.5 }{ 8.314\times { 10 }^{ -2 }\times 350 } =4.296\quad mol\)
\(\therefore\) n = 4.296 mol.

n = 5; V = 2 Iitre;T= 27 + 273 = 300 K
\(P=\frac { nRT }{ V } [\because PV=nRT]\)
\(P=\frac { 5\times 8.314\times 300 }{ 2 } =62.355\quad bar\)
\(\therefore\) The pressure of CI₂ gas will be 62.355 bar.

Given
T = 400K;\(\triangle { H }^{ 0 }\) = 77.2 kJ mol^-1 = 77200 J mol^-1 
\(\triangle { S }^{ 0 }\)=122 JK^-1 mol^-1
\(\triangle { G }^{ 0 }\) = -2.303 RT log k_eq
log k_eq =\(\frac { ​​​​​​​​​​​​​​​​​​\triangle { G }^{ 0 } }{ 2.303RT } \)
log k_eq = \(\frac { ​​​​​​​​​​​​​​​​​​\left( \triangle H^{ 0 }-T\triangle S^{ 0 } \right) }{ 2.303RT } \)
log k_eq = -\(\left( \frac { 77200-400\times 122 }{ 2.303\times 8.314\times 400 } \right) \)
log k_eq = \(-\left( \frac { 28400 }{ 7659 } \right)\)
log k_eq =-3.7080
K_eg  =antilog(-3.7080)
K_eg =1.95x10^-4

ΔH =30.56 kJ mol^-1
= 30560 J mol^-1
ΔS=6.66 \(\times\) 10^-3 kJK^-1 mol^-1
T=? at which ΔG=0
ΔG=ΔH-TΔS
0=ΔH-TΔS
T=\(\frac { \Delta H }{ \Delta S } \)
T=\(\frac { 30.56kJ\quad mol^{ -1 } }{ 6.66\times { 10 }^{ -3 }kJK^{ -1 }mol^{ -1 } } \)
T = 4589K
(i) At 4589K; ΔG= 0 the reaction is in equilibrium.
(ii) at temperature below 4598 K, ΔH>TΔS
ΔG=ΔH- TΔS > 0, the reaction in the forward direction, is non-spontaneous. In other words the reaction occurs in the backward direction.

For ethanol:
Given:
T_b =78.4°C = (78.4 + 273)
=351.4 K
\(\triangle \)H_V(ethanol) =+ 42.4 kJ mol^-1
ΔS_V =\(\frac { \triangle { H }_{ V } }{ { T }_{ b } } \)
ΔS_V  =\(\frac { +42.4KJ{ mol }^{ -1 } }{ 351.4k } \)
ΔS_V =\(\frac { +42400J{ mol }^{ -1 } }{ 351.4k } \)
ΔS_V ​​​​​​​=+120.66JK^-1mol^-1
For Toluene:
Given:
T_b =110.6°C = (110.6 + 273)
      =383.6 K
\(\triangle \)H_V (toluene)=+35.2KJ mol^-1
ΔS_V ​​​​​​​=\(\frac { \triangle Hv }{ { T }_{ b } } \)
ΔS_V ​​​​​​​=\(\frac { +35200J{ mol }^{ -1 } }{ 383.6k } \)
ΔS_V ​​​​​​​=+91.76JK^-1mol^-1

C₃H₈+5O₂\(\rightarrow \)3CO₂+4H₂O
\(\triangle { H }_{ C }^{ 0 }=-2220.2KJ\quad mo{ l }^{ -1 }\)....(1)
C+O ₂\(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }=-393.5KJ\quad mo{ l }^{ -1 }\)....(2)
\({ H }_{ 2 }+\frac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\)
\(\triangle { H }_{F}^{0 }=-285.8KJ\quad mo{ l }^{ -1 }\)...(3)
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{ C }^{ 0 }\)=?
(2) X3 \(\Rightarrow \)3C+3O₂  \(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }= \) -1180.5 KJ ....(4)
(3)X4\(\Rightarrow \) 4H₂ +2O₂ \(\rightarrow \)4H₂O
\(\triangle { H }_{F}^{0 }= \) -1143.2KJ ....(5)
(4)+(5)-(1)\(\Rightarrow \) 3C+3O₂+4H₂+2O₂+3CO₂ +4H2O\(\rightarrow \)3CO₂+4H₂O+C₃H₈+5O₂
\(\triangle { H }_{F}^{0 }= \) -1180.5-1143.2-(-2220.2)KJ
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{F}^{0 }= \) -103.5KJ
Standard heat of formation of propane is
\(\triangle { H }_{ R }^{ 0 }\left( { C }_{ 3 }{ H }_{ 8 } \right) =-103.5KJ\)

T_sys=77°C = (77 + 273) = 350 K
T_surr=33°C = (33 + 273) = 306 K
q=245 J
ΔS_sys=\(\frac{q}{T_{sys}}=\frac{-245}{350}=-0.7JK^{-1}\)
ΔS_surr=\(\frac{q}{T_{sys}}=\frac{+245}{350}=0.8JK^{-1}\)
ΔS_univ=ΔS_sys+ΔS_surr
ΔS_univ=-0.7 JK^-1+ 0.8 JK^-1
ΔS_univ=0.1 JK^-1.

n = 2 moles
V_i=500 ml = 0.5 L
V_f = 2 L
T=25⁰C = 298 K
w=-2.303 nRTlog\((\frac{V_f}{V_i})\)
w=-2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) log\((\frac{2}{0.5})\)
w -2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) log(4)
w=-2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) 0.6021
w=-6871 J
w=-6.871 KJ.

(i) Free energy is defined as G = H - TS. 'G' is a state function.
(ii) G- Extensive property; ΔG - intensive property. When mass remains constant between initial and final states of system.
(iii) 'G' has a single value for the thermodynamic state of the system.
(iv) G and ΔG values correspond to the system only.

(v) Gibbs free energy and the net work done by the system:
For any system at constant pressure and temperature
ΔG = ΔH - TΔS .....(1)
We know that,
ΔH = ΔU + PΔV
ΔG =ΔU + PΔV-TΔS
from first law of thermodynamics
ΔU = q +w
from second law of thermodynamics
Δ S=\(\frac{q}{T}\) Δ G=q+w+PΔ V-T\((\frac{q}{T}) \)
Δ G = w+PΔV
-ΔG = -w - PΔV ......(2)
But -PΔV represents the work done due to expansion against a constant external pressure.

Let us use the Born - Haber cycle for determining the lattice enthalpy of NaCl as follows:
Since the reaction is carried out with reactants in elemental forms and products in their standard states, at 1 bar, the overall enthalpy change of the reaction is also the enthalpy of formation for NaCl. Also, the formation of NaC1 can be considered in 5 steps. The sum of the enthalpy changes of these steps is equal .to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Born-Haber cycle

Δ°H_f = heat of formation of sodium chloride 
= - 411.3 kJ mol^-1
Δ°H₁= heat of sublimation ofNa(S) = 108.7 kJ mol^-1
Δ°H₂ = ionisation energy ofNa(S) = 495.0 kJ mol^-1
Δ°H₃= dissociation energy ofCI2(S) = 244 kJ  mol^-1
Δ°H₄= Electron affinity ofCl(S) = - 349.0 kJ mol^-1
Δ°H_f = Δ°H₁+Δ°H₂+1/2Δ°H₃+Δ°H₄+Δ°H₅
Δ°H₅=(Δ°H_f )-( Δ°H₁+Δ°H₂+1/2Δ°H₃+Δ°H₄)
⇒Δ°H₅ = (-411.3)-(108.7+495.0+ 122-349)
Δ°H₅= (-411.3)-(376.7)
∴ Δ°H₅ = -788 kJ mol^-1
This negative sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gaseous ions Na⁺ and Cl^-

When the system at constant pressure undergoes changes from an initial state with H₁, U₁ and, V₁ to a final state with H₂, U₂ and V₂ the change in enthalpy ΔH, can be calculated as follows:
H = U+PV
In the initial state
H₁ = U₁ +PV₁ ........(1)
In the final state
H₂ = U₂ +PV₂ ........(2)
change in enthalpy is (2) - (1)
(H₂ - H₁) = (U₂ - U₁) + P(V₂ - V₁)
ΔH = ΔU+PΔV
As per first law of thermodynamics,
ΔV = q+w
Equation (3) becomes
ΔH = q+w+PΔV
w = -PΔV
ΔH = qp-PΔV+PΔV
ΔH = qp....(4)
q_p - is the heat absorbed at constant pressure and is p considered as heat content. Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperature and pressure with V_i and V_f as the total volumes of the reactant and product gases respectively, and n _i and n_f as the number of moles of gaseous reactants and products, then,
For reactants (initial state) :
PVi = ni RT ...(5)
For products (final state) :
PVf = nfRT .........(6)
(6) - (5)
P(Vf - Vi) = (nf - ni) RT
PΔV = Δn(g) RT ...........(7)
Substituting in (7) in (3)
ΔH = ΔV +Δn(g) RT ........(8)

(i) For understanding pressure-volume work, let us consider a cylinder which contains 'n' moles of an ideal gas fitted with a frictionless piston of cross-sectional area A. The total volume of the gas inside is V_i and pressure of the gas inside is P_int.
(ii) If the external pressure P_ext is greater than P_int the piston moves inward till the pressure inside becomes equal to P_ext. Let this change be achieved in a single step and the final volume be V_f
(iii) In this case, the work is done on the system (+w). It can be calculated as follows w=-FΔx  (1)
(iv) where dx is the distance moved by the piston during the compression and F is the force acting on the gas.

F = P_extA (2)
Substituting (2) in (1)
w = - P_ext. A. Δx
A.Δx change in volume = V_f - V_i
w = - P_ext. ( V_f - V_i) (3)
w = - P_ext. (-ΔV) (4)
=P_ext.ΔV
(vi) Since work is done on the system, it is a positive quantity.
(vii) If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work done on the gas by the relation,
\({ w }_{ rev }=\int _{ { V }_{ i } }^{ { V }_{ f } }{ { P }_{ int }dV } \)
(vii) In a compression process, Pext the external pressure is always greater than the pressure of the system.
i.e P_ext= (P_int+ dP).
(viii) In an expansion process, the external pressure is always less than the pressure of the system
i.e. P_ext= (P_int - dP).

(ix) When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from V_i to V_f the P - V plot looks like in image Work done on the gas is represented by the shaded area. In general case we can write,
(x) P_ext= (P_int + dP). Such processes are called reversible processes. For a compression process work can be related to internal pressure of the system under reversible conditions by writing equation
\({ w }_{ rev }=\int _{ { V }_{ i } }^{ { V }_{ f } }{ { P }_{ int }dV } \)
For a given system with an ideal gas
P_intV= nRT
\({ P }_{ int }=\frac { nRT }{ V } \)
\({ W }_{ rev }=\int _{ { V }_{ i } }^{ { V }_{ f } }{ \frac { nRT }{ V } } dV\)
\({ W }_{ rev }=-nRT\quad \int _{ { V }_{ i } }^{ { V }_{ f } }{ \left( \frac { dV }{ V } \right) } \)  
\({ W }_{ rev }=-nRT\quad ln\quad \left( \frac { { V }_{ f } }{ { V }_{ i } } \right) \)
\({ W }_{ rev }=-2.303\quad nRT\quad log\left( \frac { { V }_{ f } }{ { V }_{ i } } \right) \)
(xi) If V_f > V_i (expansion), the sign of work done by the process is negative.
(xii) IfV_f < V_i(compression) the sign of work done on the process is positive.

(i) Heat evolved at constant volume, is measured in a bomb calorimeter.
(ii) Apparatus setup: The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.
(iii) Experiment: A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter. A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.
(iv) Calculation: A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (ΔU)_c
The amount of heat produced in the reaction (ΔU)_c is equal to the sum of the heat absorbed by the calorimeter and water.
Heat absorbed by the calorimeter q₁ = k.ΔT
where k is a calorimeter constant equal to m_c C_c (m_c is mass of the calorimeter and C_c is heat capacity of calorimeter)
Heat absorbed by the water q₂ = m_w C_w ΔT
where m_w is molar mass of water
C_w is molar heat capacity of water (4,184 kJ K^-1mol^-1)
Therefore ΔU_c = q₁ + q₂
=k.ΔT + m _w C_w ΔT
=(k+m _w Cw) ΔT
Calorimeter constant can be determined by burning a know.n mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJmol^-1)
The enthalpy of combustion at constant pressure of the substance is calculated from the equation (7.17)
\(\Delta { H }_{ C(pressure) }^{ 0 }=\Delta { U }_{ C(vol) }^{ o }+\Delta { n }_{ g }RT\)

Characteristics of internal energy (U) :

• The internal energy of a system is an extensive property.
• The internal energy of a system is a state function.
• The change in internal energy of a system is expressed as ΔU= U_f - U_i
• In a cyclic process,' there is no internal energy change. ΔU_(cyclic)= 0
• If the internal energy of the system in the final state (U_f) is less than the internal energy of the system in its initial state (U_i), then ΔU would be negative.
• ΔU = U_f- Ui = -ve (U_f < U_i)
• If the internal energy of the system in the final state (U_f) is greater than the internal energy of the system in its initial state (U_i), then ΔU would be positive.
• ΔU = U_f - U_i = +ve (U_f > U_i)

Born - Haber cycle for the formation of CaCl₂
Born - Haber cycle is used to calculate the lattice enthalpy of CaCl₂ 

Δ^oH₁ - Enthalpy change for the sublimation of Ca_(s) to Ca_(g)
Δ^oH₂ - Enthalpy change for dissociation of Cl_2(g) to 2Cl_(g)
Δ^oH₃ - Ionisation energy for Ca_(g) to Ca²⁺_(g)
Δ^oH₄ - Electron affinity for the conversion of 2Cl_(g) to 2Cl^-_(g)
Δ^oH₅ - Lattice enthalpy for the formation of solid CaCl₂·
Δ^oH_f =Δ^oH₁+Δ^oH₂+Δ^oH₃+Δ^oH₄+Δ^oH₅

Potential difference = 100V
= 100 x 1.6 x 10-¹⁹J
\(\lambda=\frac{h}{\sqrt{2mev}}\)
\(=\frac { 6.626\times { 10 }^{ -34 }Kg{ m }^{ 2 }{ s }^{ -1 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }Kg\times 100\times 1.6\times { 10 }^{ -19 }J } } \)
\(\lambda=1.22\times10^{-10}m\)

m 160 g = 160 x 10^-3 kg
\(v=140 km hr^{-1}=\frac{104\times10^{3}}{60\times60}ms^{-1}\)
\(v=38.88ms^{-1}\)
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times^{-34}Kgm^{2}s^{-1}}{160\times10^{-3}Kg\times38.88ms^{-1}}\)
\(\lambda=1.065\times10^{-34}m\)

1. This group contains Be, Mg, Ca, Sr, Ba & Ra.
2. Except Be all these elements are called as alkaline earth metals because their oxides and hydroxides are alkaline in nature.
3. Beryllium is the rare element and Radium is the rarest (10% rocks) ; Their Occurrence: Be- Beryl; Mg - carnallite; Dolomite Ca - Fluorapatite; Sr - Celestite Ba - Barytes.
4. Radium is radioactive.
5. The general electronic configuration is :
[Noble gas] ns² eg. Be - [He] 2s²
6. On moving down the group the radii increase. Their atomic radii are smaller than alkali metals.
7. They exhibit +2 oxidation state.
8. The ionisation enthalpies are less than p - block elements due to large size. Down the group the ionisation enthalpy decreases.
9. The IE₁ of group 2 elements are greater than group 1 elements.
10. The IE₂ values are higher than that of alkali metals.
11. They are less electro Positive elements than alkali metals.
12. The hydration enthalpy decreases with increase in ionic radii.
13. MgCl₂ form MgCI₂ .6H₂O and CaCl₂ form CaCl₂.6H₂O
14. The electronegativity value decreases down the group.
15. With concentrated HCl They impart flame colour. Ca - Brick Red ; Sr - Crimson red and Barium - Apple Green.
16. All form metallic halides at elevated temperatures. M + X₂ + MX₂
17. All elements except Beryllium combine with hydrogen to form hydrides of formula MH₂.

(i) Alkaline earth metal (A) belonging to 3rd period is magnesium.
(ii) So A is Magnesium. Magnesium reacts with oxygen and nitrogen as follows.
\(2Mg+O_2⟶\underset{(B)}{2MgO}\)
\(3Mg+N_2⟶\underset{(C)}{Mg_3N_2}\)
So B is Magnesium oxide and C is magnesium nitride.
(iii) Magnesium undergoes metal displacement reaction with AgNO₃ as follow to give D as follows :
\(Mg+2AgNO_3⟶\underset{D}{Mg(NO_3)_2}+2Ag\)
So D is Magnesium nitrate.
Result  :

(i) Variation of Electronegativity in a period: The electronegativity generally increases across a period from left to right. The atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.
(ii) Variation of Electronegativity in a group: The electronegativity generally decreases down a group. As we move down a group the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence, the electronegativity decreases.

Screening effect: The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where E_AB' EAA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively. The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

(i) Electron configuration of nitrogen
(Z = 7) 1s² 2s² 2p³.
Electron configuration of oxygen
(2= 8) 1s² 2s² 2p⁴.
Nitrogen has a half filled electronic configuration which is much more stable than an incomplete p-orbital of oxygen which would need to give up one of it's electrons to attain the stability of nitrogen. Hence nitrogen would require more ionization energy to remove an electron from it's outer shell than oxygen.
(ii) Electron configuration of carbon
(Z = 6) 1s²2s²2p².
Electron configuration of Boron
(Z = 5) Is²2s²2p¹
The size of a carbon atom is smaller than boron So the valence electron of carbon has greater nuclear charge than that of boron. Hence the first I.E of carbon is greater than that of boron. However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing electron, Boron has a fully filled orbital (2s²) than carbon (2p¹). Fully filled orbitals have more stability than partially filled orbitals so greater amount of energy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.
(iii) The electron affinities of Be, Mg and noble gases are almost zero because both Be (Z = 4; 1s²2s²) and Mg (Z = 12; Is²2s²2p⁶3s²) are having s orbital fully filled in their valence shell. Fully filled orbitals are most stable due to symmetry. Therefore, these elements would be having least tendency to accept electron. Hence, Be and Mg would be having zero electron affinity. Whereas N (Z = 7; 1s²2s²2px¹2py¹2pz¹ and P (Z = 15) Is² 2s² 2p⁶ 3s² 3p³ is having half filled 2p-subshell. Half filled sub shells are most stable due to symmetry (Hund's rule). Thus, nitrogen and phosphorous are having least tendency to accept electron. Hence, have low electron affinity.
(iv) Fluorine is highly electro negative in nature therefore as it gains the electron its octet become stable and releases the energy so exothermic. while in oxygen the addition of first electron is exothermic in nature but addition of second electron experiences high repulsive force. So needs extra external energy to enter outer shell, hence endothermic in nature.

r(K⁺)+r(Cl^-) = d(K^+-Cl-) = 3.14 Å.
The effective nuclear charge for K+ and CI- can be calculated as follows.
K+ = (1s²)          (2s²2p⁶)        (3s²3p⁶)
        inner shell    (n-1)th shell   n^th shell
Z*(K-) = Z-S
= 19 - [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 - 11.25 = 7.75
Z*(Cl-) = 17- [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17-11.25 = 5.75
∴ \(\frac { r({ K }^{ + }) }{ r(Cl^{ - }) } =\frac { Z*(Cl^{ - }) }{ Z*(K^{ + }) } =\frac { 5.75 }{ 7.75 } \)=0.74
∴ r(K⁺) = 0.74 r(Cl^-)
Substitute (2) in (1)
0.74 r(Cl^-) + r(Cl^-) = 3.14 Å.
1.74 r(Cl^-) = 3.14 Å
r(Cl^-) = \(\frac { 3.14\overset { 0 }{ A } }{ 1.74 } \)=1.81.Å.

The element which occupies group number (16) and period number (2) is oxygen. (B) is D₂O which is used as a moderator in nuclear reactions.
So (A) must be deuterium, which is an isotope of hydrogen
    \(2\underset { (A) }{ { D }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (B) }{ { D }_{ 2 }O } \)
So (B) is D₂O
(A) adds to (C) as follows :
\(3 \mathrm{D}_{2}+\mathrm{C}_{3} \mathrm{H}_{6} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2}\)
So (D) is 1,2 - dideutero propane. 

The group metal - 1 which is present in common salt is sodium.
So (A) is sodium
Sodium reacts with hydrogen (B) to give, sodium hydride (C). In sodium hydride the hydrogen is present in -1 oxidation state.
\(2\underset { (A) }{ Na } +\underset { (B) }{ { H }_{ 2 } } \rightarrow 2\underset { (C) }{ NaH } \)
So (B) hydrogen and (C) is sodium hydride
H₂ reacts with oxygen gas (D) to give an universal solvent, water (E) follows:
\(2\underset { (B) }{ { H }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (D) }{ { H }_{ 2 }O } \)
So (E) is water. Water is the universal solvent
Water (E)  reacts with sodium  (A) follow to give (F), which is a strong base.
\(2\underset { (E) }{ { H }_{ 2 }O } +\underset { (A) }{ 2Na } \rightarrow 2\underset { (F) }{ NaOH } +{ { H }_{ 2 } } \)
So (F) is sodium hydroxide.

The van der Waals equation for n moles is
\(\left( P+\frac { { an }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-nb \right) =nRT\) ....(1)
For 1 mole
\(\left( P+\frac { { a }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-b \right) =RT\) ...(2)
From the equation we can derive the values of critical constants P_c, V_c and T_c, in terms of a and b, the van der Waals constants, On expanding the above equation
\(pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT=0\)    ...(3)
Multiply equation (3) by V² / P
\(\frac { { v }^{ 2 } }{ p } \left( Pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT \right) =0\)
\({ v }^{ 3 }+\frac { av }{ p } +-{ bv }^{ 2 }-\frac { ab }{ { v }^{ 2 } } -\frac { { RTV }^{ 2 } }{ p } \)    ...(4)
When the above equation is rearranged in powers of Y.
\({ v }^{ 3 }-\left[ \frac { RT }{ P } +b \right] { v }^{ 2 }+\left[ \frac { a }{ p } \right] v-\left[ \frac { ab }{ p } \right] =0\)        ...(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_C. The pressure and temperature becomes Pc and T_c respectively
i.e., V = V_c
V - V_c = 0
(V - V_C)³ = 0
V3 - 3V_CV² + 3V_c²V - V_c³ = 0 .....(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V², V and constant terms in (5) and (6).
\(-3{ v }_{ c }{ v }^{ 2 }=-\left[ \frac { { RT }_{ c } }{ { p }_{ c } } +b \right] { v }^{ 2 }\)
\(3{ v }_{ c }=\frac { { RT }_{ c } }{ { p }_{ c } } +b\)           .....(7)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)           ......(8)
\(3{ v }_{ c }^{ 2 }=\frac { ab }{ { p }_{ c } } \)           ...(9)
Divide equation (9) by equation (8)
\(\frac { { v }_{ c }^{ 3 } }{ 3{ v }_{ c }^{ 2 } } =\frac { ab/{ p }_{ c } }{ a/{ p }_{ c } } \)
\(\frac { { V }_{ c } }{ 3 } =b\)
i.e. V_c = 3b          .....(10)
when equation (10) is substituted in (8)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)
\({ p }_{ c }=\frac { a }{ 3{ v }_{ c }^{ 2 } } =\frac { a }{ 3\left( { 3b }^{ 2 } \right) } =\frac { a }{ { 3\times 9b }^{ 2 } } =\frac { a }{ { 27 }b^{ 2 } } \)
\({ p }_{ c }=\frac { a }{ { 27 }b^{ 2 } } \)            ...(11)
substituting the values of V_c and P_c in equation (7),
\(3vc=b+\frac { { RT }_{ c } }{ p } \)
\(3\left( 3b \right) =b+\frac { { RT }_{ c } }{ \left( \frac { a }{ { 27b }^{ 2 } } \right) } \)
\(9b-b=\left( \frac { { RT }_{ c } }{ a } \right) 27{ b }^{ 2 }\)
\(8b=\frac { { t }_{ c }R{ 27b }^{ 2 } }{ a } \)
\(\therefore { T }_{ c }=\frac { 8ab }{ 27R{ b }^{ 2 } } =\frac { 8a }{ 27Rb } \)
\({ T }_{ c }=\frac { { 8 }_{ a } }{ 27Rb } \)           ......(12)
The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.
\(a=3{ V }_{ C }^{ 2 }{ P }_{ C }\quad and\quad b=\frac { { V }_{ C } }{ 3 } \)

The van der equation for a real gas is
\(\left( P+{{{am}^{2}}\over{{V}^{2}}} \right)(V-nb)=nRT\)
Pressure, Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
\(P^{\prime} \propto \rho^{2} ; \quad \rho=\frac{n}{v}\)
where n is the number of moles of gas and
V is the volume of the container
\( \Rightarrow p^{\prime} \alpha \frac{n^{2}}{V^{2}} \)
\(\Rightarrow p^{\prime}=a \frac{n^{2}}{V^{2}}\)
where a is proportionality constant and depends on the nature of gas
Therefore \(P_{\text {ideal }}=P+\frac{\operatorname{an}^{2}}{V^{2}}\)

Volume Correction
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container,
V. Van der Waals introduced a correction factor V' to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules
\(=\frac{4}{3} \pi(2 r)^{3}=8\left(\frac{4}{3} \pi r^{3}\right)=8 V_{m}\)
Where V_m it a volume of a single molecule
Excluded volume for single molecule = \(\frac{8 \mathrm{~V}_{\mathrm{m}}}{2}=4 \mathrm{~V}_{\mathrm{m}}\)
Excluded volume for n molecule = n(4V_m) = nb
Where b is van der waals constant which is equal to 4V_m
\( \Rightarrow V^{\prime}=n b \)
\(V_{\text {ideal }}=V-n b\)
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the Van der Waals equation of state for real gases as below,
\(\left(p+\frac{a^{2}}{V^{2}}\right)(V-n b)=n R T\)
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

The electronic configuration of Sodium (Z = 11) Is²2s²2p⁶3s¹.
Magnesium (Z = 12) 1s²2s²2p⁶3s²
Magnesium atom has a smaller radius and higher nuclear charge than a sodium atom, thus more energy will be required to remove the electron from the same orbital (3s), making the first ionisation energy of magnesium higher than that of sodium.
However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1 electron, sodium attains the stable noble gas configuration of neon (1s²2s²2p⁶). On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s²2s²2p⁶3s¹). In order to attain the stable noble gas configuration, Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

On moving diagonally across the periodic table, the second and third period elements show certain similarities. It is quite pronounced in the following pair of elements.

The similarity in properties existing between the diagonally placed elements is called diagonal relationship.

Variation along a period: Ionisation energy usually increases along a period. This is due to increase of nuclear charge and decrease in size as we move from left to right in a period.
Periodic variation in group: Ionisation energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases and hence ionisation energy also decreases down a group.

(i) Ionic radius of uni-univalent crystal can be calculated using Pauling's method from the inter ionic distance between the nuclei of the cation and anion.
(ii) Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other therefore,
d=rC⁺ + r_A^-   ...(1)
Where d is the distance between the centre of the nucleus of cation C+ and anion A-and rC⁺, rA^- are the radius of the cation and anion respectively.
(iii) Pauling also assumed that the radius of the ion having noble gas electronic configuration is inversely proportional to. the effective nuclear charge.
\({ r }_{ C }^{ + }\alpha \frac { 1 }{ ({ Z }_{ eff }){ C }^{ + } } \) ....(2) and
\({ r }_{ A }^{ - }\alpha \frac { 1 }{ ({ Z }_{ eff }){ A }^{ - } } \)...(3)
Where Z_eff is the effective nuclear charge and Z_eff= Z - S
Dividing the equation 1 by 3
\(\frac { { r }_{ C }^{ + } }{ { r }_{ A }^{ - } } =\frac { ({ Z }_{ eff }){ A }^{ - } }{ ({ Z }_{ eff }){ C }^{ + } } \) ...(4)
On solving equation and (1) and (4) the values of r_C+ and r_A^- can be obtained.

Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
\(\therefore \) When cooling starts from 700 K, H₂O will liquefy first, then followed by ammonia and finally carbon dioxide will liquefy.

V₁ = 7.05 dm³ V₂ = 2.35 dm³
n₁ = 0.312 mol n₂ = ?
' P ' and ' T ' are constant
\(\frac { { V }_{ 1 } }{ n_{ 1 } } =\frac { { V }_{ 2 } }{ { n }_{ 2 } } \)
n₂ = \(\left( \frac { { n }_{ 2 } }{ { V }_{ 1 } } \right) \times { V }_{ 2 }\)
\(\mathrm{n}_{2}=\frac{0.312 \mathrm{~mol}}{7.05 \mathrm{\not dm}^{3}} \times 2.35 \mathrm{\not dm}^{3}\)
n₂ = 0.104 mol
Number of moles exhaled = 0.312 - 0.104  = 0.208 moles

V₁ = 0.375 dm³ V₂ = 0.125
P₁ = 1.05 atm  P₂ = ?
' T '- Constant
P₁V₁  = P₂V₂
\(\therefore \) P₂  = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } =\frac { 1.05\times 0.375 }{ 0.125 } \)  = 3.15 atm

Volume of freon (V₁)  = 1. 5 dm³
Pressure  (P₁) = 0. 3atm
' T '  is constant
P₂ = 1.2 atm
V₂ = ?
\(\therefore \) P₁V₁ = P₂V₂ 
V₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { P }_{ 2 } } \)
= 
\(\therefore \) Volume decreased from 1.5 dm³ to 0.375 dm³

ΔH = -10 KJ mol^-1 = - 10000 J mol^-1
ΔS = -20 JK^-1 mol^-1
T = 300 K
ΔG = ?
ΔG = ΔH - TΔS
ΔG = -10 KJ mol^-1 - 300 k x (-20 x 10^-3 )kJ K^-1 mol^-1
ΔG = (-10 + 6) KJ mol^-1
ΔG = -4 KJ mol^-1
At 600 K
ΔG = -10 KJ mol^-1 - 600 K x (-20 x 10^-3) KJ K^-1 mol-1
ΔG = (-10 + 12) KJ mol^-1
ΔG = + 2 KJ mol^-1
The value of ΔG is negative at 300K and the reaction is spontaneous, but at 600K the value ΔG becomes positive and the reaction is non spontaneous.

T_b = 351 K
ΔH_vap = 39840 Jmol^-1
ΔS_V = ?
ΔS_v = \(\frac { { \triangle H }_{ vap } }{ { T }_{ b } } \)
ΔS_v = \(\frac { 39840 }{ 351 } \)
ΔS_v = 113.5 JK^-1 mol^-1

T₁ = 8°C + 273 = 281 K
P₁ = 6.4 atm V₁ = 2.1 ml
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm V₂ =  ?
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}} \Rightarrow \mathrm{V}_{2}=\left(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\right) \times \frac{\mathrm{T}_{2}}{\mathrm{P}_{2}}\)

= 14.25 ml

T=298K
\(\Delta { G }_{ r }^{ 0 }\)=- 13.6 kJ mol^-1
=-13600 J mol^-1
ΔG⁰=- 2.303 RT log K_eq
log K_eq=\(\frac { 13.6kJmol^{ -1 } }{ 2.303\times 8.314\times { 10 }^{ -3 }JK^{ -1 }mol^{ -1 }\times 298K } \)
log K_eq = 2.38
K_eq = antilog (2.38)
K_eq = 239.88.

T =298 K
ΔH 400 J mol^-1 = 400 J mol^-1
ΔS= 0.2 J K^-1 mol^-1
ΔG =ΔH-TΔS
if T =2000 K
ΔG = 400 - (0.2 \(\times\) 2000) = 0
if T > 2000 K
ΔG will be negative.
The reaction would be spontaneous only beyond 2000K.

\(\Delta { H }_{ f }^{ 0 }\)(SO₂) = - 297 kJ mol^-1
\(\Delta { H }_{ f }^{ 0 }\)(SO₃) = - 396 kJ mol^-1
SO₂ + \(\frac{1}{2}\)O₂⟶SO₃ \(\Delta { H }_{ r}^{ 0 }\)=?
\(\Delta { H }_{ r}^{ 0 }\)=\((\Delta { H }_{ f }^{ 0 })\)_compound-Σ(ΔH_f)_elements
\(\Delta { H }_{ r}^{ 0 }\)=​​​​\(\Delta { H }_{ f }^{ 0 }\)(SO₃)-(\(\Delta { H }_{ f }^{ 0 }\)(SO₂) +\(\frac{1}{2}\)\((\Delta { H }_{ f }^{ 0 })\)(O₂))
\(\Delta { H }_{ r}^{ 0 }\)=- 396 kJ mol^-1 - (-297 kJ mol^-1 + 0)
\(\Delta { H }_{ r}^{ 0 }\)=-396 kJ mol^-1 + 297
\(\Delta { H }_{ r}^{ 0 }\)=-99 kJ mol^-1

T= 33K
N₂O₄ \(\rightleftharpoons \) 2NO₂
Initial concentration 100% 
Concentration dissociated 50% 
Concentration remaining at equilibrium 50% 100%
K_eq=\(\frac{100}{50}=2\)
ΔG^o=-2.303 RT log Keq
ΔG^o=-2.303 x 8.314 x 33 x log 2
ΔG^o=-190.18 J mol^-1

E_C-H=414 kJ mol^-1
E_C-C=347 kJ mol^-1
E_C=C=618 kJ mol^-1
E_H-H=435 kJ mol^-1

ΔH_r= Σ(Bond energy)_r -Σ(Bond energy)p
ΔH_r= (E_C=C + 4E_C-H + E_H-H) - (E_C-C + 6E_C-H)
ΔH_r= (618 + (4 \(\times\) 414) + 435) - (347 +(6 \(\times\) 414))
ΔH_r= 2709 - 2831
ΔH_r​​​​​​​= -122 kJ mol^-1

T = 298K ; ΔU= - 742.4 kJ mol^-1
ΔH=?
ΔH=ΔU+Δn_gRT
ΔH=ΔU+(n_p-n_r)RT
ΔH=-742.4 +\((2-\frac{3}{2})\)\(\times\)8.314 \(\times\) 10^-3 \(\times\) 298
=-742.4 + (0.5 \(\times\) 8.314 \(\times\)10^-3 \(\times\)298)
=-742.4 + 1.24
=-741.16 kJ mol^-1

ΔH_f (NaCl)=30.4 kJ = 30400 J mol^-1
ΔS_f (NaCl)=28.4 JK^-1mol^-1
T_f=?
ΔS_f =\(\frac { \Delta { H }_{ f } }{ { T }_{ f } } \)
T_f=\(\frac { \Delta { H }_{ f } }{ \Delta { S }_{ f } } \)
T_f=\(\frac { 30400J{ mol }^{ -1 } }{ 28.4J{ K }^{ -1 }mol^{ -1 } } \)
T_f=1070.4K

n = 1mole
P = 4.1 atm
V= 2 L
T=?
q=3710 J
ΔS=\(\frac{q}{T}\)
ΔS=\(\frac { q }{ \left( \frac { PV }{ nR } \right) } \)
ΔS=\(\frac{nRq}{PV}\)
ΔS=\(\frac { 1\times 0.082\quad lit\quad atm{ K }^{ -1 }\times 3710J }{ 4.1\quad atm\times 2\quad lit } \)
ΔS​​​​​​​=37.10JK^-1

T_i = 298 K
T_f = 298.45 K
k = 2.5 kJK^-1
m =3.5g
M_m = 28
heat evolved= kΔT
=k (T_f - T_i)
=2.5 KJ K^-1 (298.45 - 298)K
=1.125kJ
ΔH_c=\(\frac{1.125}{3.5}\times\)28KJmol^-1
ΔH_c = 9 kJ mol^-1

\(\triangle x=?\)
\(\triangle v=?\)
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\lambda (m\triangle v)\ge\frac{h}{4\pi}\)
\(\triangle v\ge\frac{h}{4\pi(m\lambda)}\)
\(\triangle \ge\frac{h}{4\pi\times m \times\frac{h}{mv}}\)   \(\left[ \because \lambda-={{h}\over{mv}} \right]\)
\(\triangle v \ge \frac{v}{4\pi}\)
therefore, minimum uncertainty in velocity \(=\frac{v}{4\pi}\)

\(\triangle\)x = 0.6\(\mathring{A}\) = 0.6 x 10^-10m
\(\triangle p\) = ?
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\triangle x.\triangle p\ge5.28\times10^{-35}kgm^{2}s^{-1}\)
\((0.6 \times10^{-10}) \triangle p \ge5.28\times10^{-35}\)
\(\Rightarrow \triangle \ge \frac{5.28\times10^{-35}kgm^{2}s^{-1}}{0.6510^{-1}m}\)
\(\triangle p \ge8.8\times10^{-25}kgms^{-1}\)

v = 2.85 x 10⁸ ms^-1
m_p = 1.673 x 10^-27Kg
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}kgm^{2}s^{-1}}{1.673\times10^{-27}kg\times\times2.85\times10^{8}ms^{-1}}\)
\(\lambda=1.389\times10^{-15}\Rightarrow\lambda=1.389\times10^{-15}A\) \([\because \mathring{A}={10}^{-10}m]\)

\({ r_{n} }=\frac{(0.529)n^2}{z}\mathring{A}\ \ { E_{n} }=\frac{-13.6(z)^2}{(n)^2}ev atom^{-1}\)
for Li²⁺ z = 3
Bohr radius for the third orbit (r₃)
= \(\frac { (0.529){ (3) }^{ 2 } }{ 3 } \)
= 0.529\(\times\)3
=1.587 \(\mathring{A}\)
Energy of an electron in the fourth orbit
\(({E}_{6})=\frac { -13.6{ (3) }^{ 2 } }{ { (4) }^{ 2 } } \)
=-7.65eV atom^-1

\({ He }^{ + }\longrightarrow { He }^{ 2+ }+{ e }^{ - }\)
\({ E }_{ n }=\frac{-13.6(2)^2}{n^2}\)
\({ E }_{ 1 }=\frac{-13.6(2)^2}{(1)^2}=-54.4\)
\({ E }_{ \infty }=\frac{-13.6(2)^2}{(\infty)^2}=0\)
\(\therefore\) Required Energy for the given process
= E_\(\infty\) - E_I = 0 - (- 54.4) = 56.4 ev.

Pressure of the gas in the tank at its melting point
T₁ = 298 K; P₁ = 2.98 atm; T₂ = 1100 K; P₂ = ?
\(\frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow \quad { P }_{ 2 }=\frac { { P }_{ 1 } }{ { T }_{ 1 } } \times { T }_{ 2 }\)
\({ P }_{ 2 }=\frac { 2.98\quad atm }{ 298\ K } \times 1100\ K=11\ atm\)
At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first. 

\(\frac { { \gamma }_{ unknown } }{ \gamma { N }_{ 2 } } =\frac { { t }_{ { N }_{ 2 } } }{ { t }_{ unknown } } =\sqrt { \frac { { m }_{ { N }_{ 2 } } }{ { m }_{ unknown } } } \)
\(\frac { 84\quad sec }{ 192\quad sec } =\sqrt { \frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } } =\frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } \)
\({ m }_{ unknown }=28 g { mol }^{ -1 }\times { \left( \frac { 192\quad sec }{ 84\quad sec } \right) }^{ 2 }\)
m_unknown = 146 g mol^-1 . 

Given,
V = 154.4 \(\times\) 10^-3 dm³,
P = 742 mm of Hg
T = 298 K m = ?
\(n=\frac { PV }{ RT } =\frac { 742mm\ Hg\times 154.4\times { 10 }^{ -3 }L }{ 62mm\ Hg\ L{ K }^{ -1 }{ mol }^{ -1 }\times 298K } \)
\(n=\frac { Mass }{ Molar\ Mass } \)
mass = n \(\times\) Molar mass
= 0.006 \(\times\) 2.016
= 0.0121 g = 12.1 mg.

T₁ = 6°C + 273 = 279 K
P₁ = 4 atm V₁ = 1.5 ml
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm V₂= ?
\(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \)

V₂ = 6.41 ml.

Given that, y = x + 11.1% of x
\(=\left(x+\frac{11.1}{100} x\right)=x+0.111 x\)
Y = 1.111 x
mass number = 37
number of protons + number of neutrons = 37
(x -1) + 1.111x  = 37
x+1.111x = 38
2.111x  = 38
\(x=\frac{38}{2.11}\)
x = 18.009
x 18 (whole number)
ஃ  Atomic number = x- 1 = 18 - 1 = 17
Mass number = 37
Symbol of the ion \({ _{ 17 }^{ 37 }{ Cl } }^{ - }\)

V₁ = 8.5 dm³ V₂ = 6.37 dm³
T₁ = ? T₂ = 0^o C = 273 K
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 1 }\times \left( \frac { { T }_{ 2 } }{ { V }_{ 2 } } \right) ={ T }_{ 1 }\)

T₁ = 364.28 K

T₁ = 15^oC + 273 T₂= 38 + 273
T₁ = 288 K  T₂ = 311 K
V₁ = 2.58 dm³ V₂ = ?
(P = 1 atm constant)
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\left( \frac { { V }_{ 1 } }{ { T }_{ 1 } } \right) \times { T }_{ 2 }\)

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
 At anode: 2OH^- ➝ H₂O + 1/2O₂ + 2e^-
At cathode: 2H₂O + 2e^- ➝ 2OH^- + H₂
Overall reaction: H₂O ➝ H₂ + 1/2O₂ 

(i) Hydrogen has the electronic configuration of 1s1 which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals (Na⁺,K⁺,Cs⁺)
2. It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
3. It also acts as a reducing agent.
4. lt is an electro positive element
However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1.314 kJ mol^-1 which is much higher than alkali metals.
Like the formation of halides (X ^-) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion(H⁺) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\( 1 / 2 \mathrm{H}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{H}^{-} \Delta \mathrm{H}=+36 \mathrm{kcal} \mathrm{mol}^{-1} \)
\(1 / 2 \mathrm{Br}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-} \Delta \mathrm{H}=-55 \mathrm{kcal} \mathrm{mol}^{-1}\)
Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Aerated water bottle contains excess dissolved oxygen/CO₂ bottled at high pressure. During summer due to high Temperature, O₂/CO₂ escapes from the solution increasing the Pressure inside the bottle which may cause bursting of the bottle. To maintain low T, it is kept under water

Given \(\triangle\)v = 5.7 x 10⁵ ms^-1. \(\triangle\)x = ?
According to Heisenbergs uncertainty principle \(\Delta x \cdot \Delta p \geq \frac{\mathrm{h}}{4 \pi}\)
\( \frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14} \mathrm{kgm}^{2} \mathrm{~s}^{-1}=5.28 \times 10^{-35} \)
\(\Delta x \cdot \Delta \mathrm{p} \geq 5.28 \times 10^{-35} \)
\(\Delta x . \mathrm{m} \Delta \mathrm{v} \geq 5.28 \times 10^{-35} \)
\(\Rightarrow \Delta x \geq \frac{5.28 \times 10^{-35} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 5.7 \times 10^{5} \mathrm{~ms}^{-1}} \Rightarrow \Delta x \geq 1.017 \times 10^{-10} \mathrm{~m}\)

No. of moles of Oxygen =\(\frac { Mass }{ Molar Mass }\) =\(\frac { 52.5}{ 32 }\)
= 1.640 moles
No. of moles of CO₂ = \(\frac { Mass }{ Molar Mass }\)=\(\frac { 65.1 }{ 44} \)
= 1.480 moles
Partial pressure = Mole fraction x Total Pressure
\(\therefore \mathrm{P}_{\mathrm{o}_{2}}=\left(\frac{1.641}{1.641+1.480}\right) 9.21=\frac{1.641}{3.121} \times 9.21\)
= 4.842 atm.
\(\mathrm{P}_{\mathrm{CO}_{2}}=\left(\frac{1.480}{1.641+1.480}\right) 9.21=\frac{1.480 \times 9.21}{3.121}\)
= 4.367 atm.

V₁ = 3.8 dm³ T₂ = 0^oC = 273 K
T₁ = ? V₂ = 2.27 dm³
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } { T }_{ 1 }=\left( \frac { { T }_{ 2 } }{ { V }_{ 2 } } \right) \times { V }_{ 1 }\)
 
T₁ = 457K

Given data P₁ = 1.2 atm
T₁ = 18° C
T₁ = 18 + 273 = 291 K
P₂ = ?
T₂ = 85° C
= 85 + 273 = 358 K
\(\frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } \)
\({ P }_{ 2 }=\frac { { P }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1.2\times 358 }{ 291 } \)
P₂ = 1.48 atm

n =1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
\(P=\frac { nRT }{ V } \)

P = 9.425 atm.

n_A = 1.5 mol n_B = ?
V_A= 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)
\(\frac { { V }_{ A } }{ { n }_{ A } } =\frac { { V }_{ B } }{ { n }_{ B } } \)
\({ n }_{ A }=\left( \frac { { n }_{ A } }{ { V }_{ A } } \right) { V }_{ B }\)

= 0.66 mol.

The gas deviates from ideal gas behaviour and will be a real gas only. In the compressed state, the inter molecular forces will be very high as the molecules are very close.