Both gases have same value of average translational kinetic energy per mole because their temperatures are equal and \(\bar { { E } } =\frac { 3 }{ 2 } RT\)

\(\frac { 3{ k }_{ B }T }{ 2 } \)

Since C\(\propto \sqrt { T } \)
Therefore, the r.m.s. velocity becomes  \(\sqrt { 3C } \)
Hence, increase in r.m.s. velocity = \(\sqrt { 3C } \) - C 
= 0.732 C

P₁/T₁=P₂/T₂ or T₂=P₂ T₂/P₁ = \(\frac { 2\times (273-173) }{ 1 } \) = 200 K =-73⁰C

The ratio of r.m.s velocities is given as
\(\frac { { { C }_{ 1 } } }{ { C }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } =\sqrt { \frac { { \rho }_{ 2 } }{ { \rho }_{ 1 } } } \)
\(\frac { { C }_{ 1 } }{ { C }_{ 2 } } =\sqrt { \frac { 9 }{ 6 } } =\sqrt { 3 } :\sqrt { 2 } \)

Given, T = 27⁰C = 273 + 27 = 300K,
p = 1atm = 1.01 x 10⁵N/m²
d = 2 x 2 x 10^-10m = 4 x 10^-10
\(\therefore \ Mean\ free\ path,\ \lambda ={ K }_{ B }T/\sqrt { 2 } \pi { d }^{ 2 }p\)
\(\\ =\frac { 1.38\times { 10 }^{ -23 }\times 300 }{ 1.414\times 3.14(4\times { 10 }^{ -10 })^{ 2 }1.013\times { 10 }^{ 5 } } =5.75\times { 10 }^{ -8 }m\)

Oxygen is a diatomic gas.
Number of moles of O₂ gas
\(=\frac { Atomic\ wt. }{ Molecular\ wt. } =\frac { 8 }{ 32 } \)
\(\\ =\frac { 1 }{ 4 } =0.25\)
\(\\ \therefore \ Energy\ associated\ with\ 1\ mole\ of\ oxygen\)
\(\\ U=\frac { 5 }{ 2 } RT\)
\(\\ \therefore \ Internal\ enreyg\ of\ 8g\ of\ oxygen=0.25\times \frac { 5 }{ 2 } \times 8.31\times 273=1417.9J\)

For 1 mole with f degrees of freedom,
Internal energy, U = 1 x C_v x T = f²/RT
For n moles, U = nC_vT = nf²/RT

Degree of freedom, f = 3N - K.

(i) We know that (KE)_trans =  \(\frac { 3 }{ 2 }\) nRT
Given, n=2 mol, T = 273 + 20 = 293 K
\(\Rightarrow\) (KE)_trans =\( \frac { 3 }{ 2 }\) (2)(8.31)(293)
= 7.3 x 10³J
(ii) Average KE per molecule
= \(\frac { 1 }{ 2 }\) \({ mv }_{ rms }^{ -2 }\)
= \(\frac { 3 }{ 2 }\) kT
=\(\frac { 3 }{ 2 } \)(1.38\(\times \)\({ 10 }^{ -23 }\))(293)
= 6.07 x 10^-21 J

The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) k_B T. It depends only on temperature, and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1.
(ii) Now ½ m v_rms² = average kinetic energy per molecule = (3/2) ) k_B T where m is the mass of a molecule of the gas. Therefore,
\( \frac{(v^2_rms)_{Ar}}{(v_rms)cl_2}\) = \( \frac{(m)cl}{(m)_{Ar}}\) =\(\frac{M_{cl}}{M_{Ar}}\) = \(\frac{70.9}{39.9}\) = 1.777
where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides,
\(\frac{(v_{rms})_{Ar}}{(v_{rms})_cl_2}\) = 1.333
You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.

As for one mole of ideal gas,  pV =\(\mu\)RT
pV = RT
V = \(\frac{RT}{p}\)
Putting , R = 8.31 Jmol^-1K^-1, T = 273 K
p = 1 atm = 1.013 \(\times \)10⁵ N -m^-2
v =  \(\frac{8.31\times273}{1.013\times10^5}\)
= 0.0224 m³
= 22.4 L  [ \(\therefore\) 1 m³ = 10³L ]

If a gas is heated, then the root mean square velocity of its molecules is increased.
∴ V_rms ∝ √T
∴ The temperature of the gas increases .

When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.

For monoatomic gas, N =1
The total degree of freedom  = 3
\( C_V=\frac{3 R}{2} \)
\( \text { Since } C_P=C_V+R \)
\( C_P=\frac{3 R}{2}+R \)
\( C_P=\frac{5 R}{2}\)
\( Y=\frac{\frac{5 R}{2}}{\frac{3 R}{2}} \)
\( Y=\frac{5}{3}=1.66 \)

According to kinetic theory of gases,
\(p=\frac { 1 }{ 3 } { P }^{ { C }^{ 2 } }=\frac { 1 }{ 3 } \frac { M }{ V } { C }^{ 2 }\)
\(=\frac { 1 }{ 3 } \frac { M }{ V } KT\)   
\( T\propto PV\)      \((\because { C }^{ 2 }=KT.\)When k is constant)
Now as T is directly proportional to the product of P and V. If P and V are constant, then T is also constant.

(i) Average speed,
\({ v }_{ av }=\frac { 0+2+3+4+4+4+5+5+6+9 }{ 10 } \)
\(=\frac { 42 }{ 10 } =4.2\quad ms^{ -1 }\)
(ii) R.M.S. Speed,
\({ v }_{ rms }=\left[ \frac { { (0) }^{ 2 }+{ (2) }^{ 2+ }{ (3) }^{ 2 }+{ (4) }^{ 2 }+{ (4) }^{ 2 }+{ (4) }^{ 2 }+{ (5) }^{ 2 }+{ (5) }^{ 2 }+{ (6) }^{ 2 }+{ (9) }^{ 2 } }{ 10 } \right] ^{ 1/2 }\)
\(=\left[ \frac { 228 }{ 10 } \right] ^{ 1/2 }=4.77\quad ms^{ -1 }\)

Temperature T = 300K
Volume V = 4 litre = 4 x 10^-3 m³ 
The pressure exerted by a gas is given by
\(p=\frac { nRT }{ V } =\frac { mass }{ molecular\ weight } \times \frac { RT }{ V } \)
Pressure exerted by oxygen P₁= \(\\ \frac { 8 }{ 32 } \frac { RT }{ V } =\frac { 1 }{ 4 } \frac { RT }{ V } \)
Pressure exerted by nitrogen P₂ = \(\frac { 14 }{ 28 } \frac { RT }{ V } =\frac { 1 }{ 2 } \frac { RT }{ V } \)
Pressure exerted by carbon dioxide P₃ = \(\\ \frac { 22 }{ 44 } \frac { RT }{ V } =\frac { 1 }{ 2 } \frac { RT }{ V } \)
From Dalton's law of partial pressures, the total pressure exerted by the mixture is given by
P = P₁ + P₂ + P₃
\(\\ \frac { 1 }{ 4 } \frac { RT }{ V } =\frac { 5 }{ 4 } \times \frac { 8.315\times 300 }{ 4\times 10^{ -3 } } \)
= 7.79 x 10⁵ N m^-2

\(For\ a\ monoatomic\ gas,\ i.e.\gamma =\frac { 5 }{ 3 }\)
\( \\ { C }_{ { V }_{ \gamma } }=\frac { R }{ \gamma -1 } =\frac { R }{ \frac { 5 }{ 3 } -1 } =\frac { 3 }{ 2 } R\)
\(\\ For\ a\ diamotic\ gas,\ i.e.\ \gamma =\frac { 7 }{ 5 }\)
\( \\ { C }_{ V }=\frac { R }{ \frac { 7 }{ 5 } -1 } =\frac { 5 }{ 2 } R\)
\(\\ By\ conservation\ of\ energy,\)
\(\\ { C }_{ { V }_{ mixture } }=\frac { { \mu }_{ 1 }{ C }_{ { V }_{ 1 } }+{ \mu }_{ 2 }{ C }_{ { V }_{ 2 } } }{ { \mu }_{ 1 }+{ \mu }_{ 2 } } \)
\(=\frac { 2\times \frac { 3 }{ 2 } R+3\times \frac { 5 }{ 2 } R }{ 2+3 } =\frac { 3R+7.5R }{ 5 } =2.1\ R\)

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P₁V=μ₁,RT and P2V=μ2RT, i.e. (P1/P2)=(μ₁/μ2). Here 1 and 2 refer to neon and oxygen respectively. Since (P₁/P₂)=(3/2) (given), (μ₁/μ₂)=3/2.
(i) By definition μ₁=(N₁/NA) and u2=(N₂/NA) where N₁ and N₂ are the number of molecules of 1 and 2, and NA is the Avogadro.s number. Therefore, (N₁/N₂)=(μ₁/μ₂)=3/2.
(ii) We can also write μ₁=(m₁/M₁) and μ₂=(m₂/M₂) where m₁ and m₂ are the masses of 1 and 2, and M₁ and M₂ are their molecular masses. (Both m₁ and M₁, as well as m₂ and M₂ should be expressed in the same units). If ρ₁ and ρ₂ are the mass densities of 1 and 2 respectively, we have
\( \frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \)   
\(=\frac{3}{2} \times \frac{20.2}{32.0}=0.947 \)

2.25 R

We know that r.m.s. speed is given by
\(\\ { v }_{ rms }=\left[ \frac { 3PV }{ M } \right] ^{ 1/2 }=\left[ \frac { 3RT }{ M } \right] ^{ 1/2 }\)
If (v_rms)l be the r.m.s. speed of oxygen and (v_rms)be the r.m.s. of helium gas at temperature T₁ and T₂ respectively.
\({ (v }_{ rms })_{ 1 }=\left[ \frac { 3RT_{ 1 } }{ { M }_{ 1 } } \right] ^{ 1/2 }\quad and\quad { (v }_{ rms })_{ 2 }=\left[ \frac { 3RT_{ 2 } }{ { M }_{ 2 } } \right] ^{ 1/2 }\)
\(or\quad \frac { { (v }_{ rms })_{ 1 } }{ { (v }_{ rms })_{ 2 } } =\left[ \frac { { M }_{ 2 }{ T }_{ 1 } }{ { M }_{ 1 }{ T }_{ 1 } } \right] ^{ 1/2 }\)
here \({ (v }_{ rms })_{ 1 }={ (v }_{ rms })_{ 2 },{ M }_{ 1 }=32,{ M }_{ 2 }=4.0;\)
T₁ = ?,T₂ = -10 + 273 = 263K
\(\\ \therefore \ 1=\left[ \frac { 4\times { T }_{ 1 } }{ 32\times 263 } \right] ^{ 1/2 }\)
\(\\ or\quad { T }_{ 1 }=\frac { 32\times 263 }{ 4 } =2104K\)

(a) The dotted plot corresponds to 'ideal' gas behaviour as it is parallel to P-axis and it tells that value of PV /T remains same even when P is changed
(b) The upper position of PV /T shows that its value is lesser for T₁, thus T₁> T_2. This is because the curve at T₁ is more close to dotted plot than the curve at T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the tempera lure is increased
(c) Where the two curves meet, the value of PV / T on y-axis is equal to \(\mu \)R Since ideal gas
equation for \(\mu \) moles is PV = \(\mu \)RT
\(where,\ \mu =\frac { 1.00\times 10^{ -3 }kg }{ 32\times 10^{ -3 }kg } =\frac { 1 }{ 32 } \)
\(\therefore \ value\ of\frac { PV }{ T } \mu R=\frac { 1 }{ 32 } \times 8.31\quad { JK }^{ -1 }\)
(d) If we obtained similar plots for 1.00 x 10^-3 kg of hydrogen, we will not get the same value of \(\frac { PV }{ T } \)  at the point, where the curves meet on the y-axis. This is because molecular mass of hydrogen is different from that of oxygen.
For the same value of \(\frac { PV }{ T } \) mass of hydrogen required is obtained from
\(\frac { PV }{ T } nR=\frac { m }{ 2.02 } \times 8.31=0.26\)
\(\\ m=\frac { 2.02\times 0.26 }{ 8.31 } gram=6.32\times { 10 }^{ -2 }gram\) 

Initially in the oxygen cylinder, V₁=30 litre =\(30 \times 10^{-3} \mathrm{~m}^3\),
\(P_1=15 a t m=15 \times 1.01 \times 10^5 \mathrm{~Pa}, T_1=27+273 =300 \mathrm{~K}\)
If the cylinder contains n₁ mole of oxygen gas, then \(P_1 V_1=n_1 R T_1\)
or
\(n_1=\frac{P_1 V_1}{R T_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300} =18.253\)
For oxygen moleculaer weight, M=32 g
Initial mass of oxygen in the cylinder cylinder,
\(m_1=n_1 M=18.253 \times 32=548.1 g\)
Finally in the oxygen gas in the cylinder, let n₂ moles of oxygen be left,
Here,
\( V_2=30 \times 10^{-3} \mathrm{~m}^3, P_2=11 \times 1.01 \times 10^5 \mathrm{~Pa}, T_2 =17+273=290 \mathrm{~K}\)
Now,
\( n_2=\frac{P_2 V_2}{R T_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290} =13.847 \)
\( \therefore \text { Final mass of oxygen gas in the cylinder, } m_2=13.847 \times 32=453.1 \mathrm{~g} \)
\( \therefore \text { Mass of the oxygen gas withdrawn }=m_1-m_2=584.1-453.1=131.0 \mathrm{~g} . \)