Between 'b' and 'c' is the portion of isotherm at which liquid CO₂ is in equilibrium with gaseous CO₂.

Total K.E = 74826 J Average K.E, K.E = 6.1 x 10^-21 J mole^-1 J

P_t = 25.21 atm

According to Boyle's law,  \({ p }_{ 1 }{ V }_{ 1 }={ p }_{ 2 }{ V }_{ 2 }\)
\(1bar\times 2.27L=0.2bar\times { V }_{ 2 }\)
\(or \ { V }_{ 2 }=\frac { 1bar\times 2.27L }{ 0.2bar }\)
\( =11.35L\)
Since, balloon burst at 0.2 bar pressure, the volume of balloon should be less than 11.35 L.

Apply ideal gas equation, pV = nRT
\(T=\frac { pV }{ Rn } =\frac { 3.32 \ bar\times 5{ dm }^{ 3 } }{ 0.083 \ bar \ { dm }^{ 3 }{ k }^{ -1 }{ mol }^{ -1 }\times 4 \ mol } \)
\(T=50 \ k \)

\({ n }_{ { N }_{ 2 } }=\frac { 14 }{ 28 } =0.05mol\)
\( 1mol=6.022\times { 10 }^{ 23 } \ molecules\)
\( 0.05mol=0.05\times 6.022\times { 10 }^{ 23 }\)
\(=0.3011\times { 10 }^{ 23 }\quad molecules\)
\( 1 \ molecules \ of \ { N }_{ 2 } \ contains \ -14 \ electrons\)
\( 0.3011\times { 10 }^{ 23 } \ 0.3011\times { 10 }^{ 23 } \ will \ contain\)
\( =0.3011\times { 10 }^{ 23 }\times 14=4.2154\times { 10 }^{ 23 } \ electrons\)

Critical temperature is the temperature above which the gas cannot liquefied, how so ever high pressure may be applied.
Higher the critical temperature, more easily the gas can be liquefied i,e stronger are the intermolecular forces.Hence,CO₂  has stronger intermolecular forces than CH₄

For ideal gas,compressibility factor, Z = 1

As the temperature of a liquid increases, kinetic energy of the molecules increases which can overcome intermolecular forces. So, the liquid can flow more easily, this results in decrease in viscosity of the liquid.

Consider two molecules to be spherical, if r is the radius of the molecules then the distance of closest approach between the two molecules = 2r (as shown in Figure). This is the distance between the centres of their nuclei.

Since the molecules cannot come closer than distance 2r, the excluded volume for a pair of molecules = volume of sphere of radius 2r
\(=\frac { 4 }{ 3 } \pi { \left( 2r \right) }^{ 3 }=8\times \frac { 4 }{ 3 } \pi { r }^{ 3 }\) 
\(\therefore \) Excluded volume per molecules (b)
\(=\frac { 1 }{ 2 } \left( 8\times \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =4\times \frac { 4 }{ 3 } \pi { r }^{ 3 }\)  
But \(\frac { 4 }{ 3 } \pi { r }^{ 3 }={ V }_{ m }\), i.e. the actual volume of the gas molecule.
\(\therefore \)  b = 4V_m

 Density, d = pM/RT
For same gas at different temperatures and pressures
\(\cfrac { { d }_{ 2 } }{ d_{ 1 } } =\cfrac { { p }_{ 2 }{ T }_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } }\)
\( { d }_{ 2 }=\cfrac { { p }_{ 2 }{ T }_{ 1 }d_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } } \)
\(\cfrac { 1\times 300\times 5.46 }{ 2\times 273 } \)
= 3 g dm^-3

Moles of O₂, ⁿO₂ = \(\cfrac{mass}{mol.wt.}=\cfrac{8}{32}=0.25mol\)
[mol.wt. of O₂ = 16\(\times\)2 = 32]
Moles of H₂, ⁿH₂ = \(\cfrac{4}{2}=2.0mol\)
[mol.wt. of H₂ = 1 \(\times\)2 = 2 ]
Total number of moles = 0.25 + 2.0 = 2.25mol
Pressure, p = \(\cfrac{nRT}{V}\)
\(\cfrac { 2.25mol\times 0.083bar \ dm^{ 3 }K^{ -1 }mol^{ -1 }\times 300K) }{ (1 \ dm^{ 3 }) } \)
p = 56.025 bar

Suppose volume of the air in flask = V cm³  at 27\(^{o}\) C (300K).
 V₁ = Vcm³ , V₂ = ?, T₁= 300 K, T₂ = 750 K
\(\cfrac { { V }_{ 1 } }{ { T }_{ 1 } } =\cfrac { { V }_{ 2 } }{ { T }_{ 2 } } \)
or \(\cfrac { { V } }{ 300 } =\cfrac { { V }_{ 2 } }{ { 750 } } \)
or   \(300{ V }_{ 2 }=750V\)
or  \({ V }_{ 2 }=2.5V\) (final volume)
Volume expelled = 2.5 V - V = 1.5 V
Fraction of air expelled = \(\cfrac{1.5 V}{2.5 V}=0.6\)