The ideal gas equation relating pressure (P),
volume (v), and absolute temperature (T) is
given as: pv = nRT, where R is the universal gas
constant = 8.314 J mol^-1 K^-1.
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1atm = 1.013\(\times\)10⁵ Nm^-2
\(\therefore v=\frac { nRT }{ P } =\frac { 1\times 8.314\times 273 }{ 1.013\times { 10 }^{ 5 } } \)
\(=\frac { 2269.7 }{ 1.013 } \times { 10 }^{ -5 }\)
= 0.0224m³
= 22.4 litres
Hence, the molar volume of gas at STP is 22.4 litres.

temperature of the helium atom,
T_He= -20^oC = 273 - 20 = 253 K.
Atomic mass of argon, M_Ar = 39.90,
Atomic mass of helium, M_He= 4.04.
Let, (V_rm) Ar be the rms speed of argon,
Let, (V_rms) He be the rms speed of Helium.
The rms speed of argon is given by:
(V_rms)_Ar=\(\sqrt { \frac { { 3RT }_{ AR } }{ { M }_{ Ar } } } \)  ............ (1)
(V_rms)_He=\(\sqrt { \frac { { 3RT }_{ He } }{ { M }_{ He } } } \) ............ (2)
It is given that:
(V_rms) _Ar = (V_rms)_He
\(\sqrt { \frac { { 3RT }_{ AR } }{ { M }_{ Ar } } } \)=\(\sqrt { \frac { { 3RT }_{ He } }{ { M }_{ He } } } \)
\(\frac { { T }_{ AR } }{ { M }_{ Ar } } =\frac { { T }_{ He } }{ { M }_{ He } } \)
\({ T }_{ Ar }=\frac { { T }_{ He } }{ { M }_{ He } } \times { M }_{ Ar }\)
\(=\frac { 253 }{ 4 } \times 39.9\)
= 2523.675 = 2.52\(\times\)10³K.
Therefore, the temperature of the argon atom is 2.52\(\times\)10³ K.

Number of molecule in 0.014 kg of nitrogen
\(n=\frac { 0.014\times { 10 }^{ 3 } }{ 27 } \)=0.0005\(\times\)103
\(=0.5=\frac { 1 }{ 2 } \)mole.
Molecular specific heat
\({ C }_{ V }=\frac { 5 }{ 2 } \times R\)
\({ C }_{ V }=\frac { 5 }{ 2 } R=\frac { 5 }{ 2 } \times 2\)
\(=\frac { 10 }{ 2 } =5\)cal / molecule k
[\(\therefore\) Nitrogen diatom molecule S_o=\(\frac{5}{2}\)]
Ratio of volume  \(\frac { { V }_{ 2 } }{ { V }_{ 1 } } =\sqrt { \frac { { T }_{ 2 } }{ { T }_{ 1 } } , } { T }_{ 2 }=4{ T }_{ 1 }\)
\(\Delta\) T=T₂-T₁=4T₁-T₁=3T₁
= 3\(\times\)300=900K
\(\Delta\)Q = nC_V\(\Delta\)T= \(\frac{1}{2}\)\(\times\)5\(\times\)900
\(\therefore\)Heat transfer \(\Delta\)Q = 2250 cal

For monoatomic gas, C_v = \(\frac{3}{2}\)R, n = 1molecule
For diatomic gas, C_v = \(\frac{3}{2}\)R, n = 1molecule
From conservation of energy, the molecular specific heat of the mixture is,
\({ C }_{ v }=\frac { n\left( { C }_{ v } \right) +n'\left( { C }_{ v }' \right) }{ n+n' } \)
\(=\frac { 1\times \frac { 3 }{ 2 } R+3\times \frac { 5 }{ 2 } R }{ (2+3) } \)
\(=\frac { 1.5R+7.5R }{ 4 } \)
\(=\frac { 9R }{ 4 } \)
\(\therefore\) C_v'= 9\(\times\)8.31 = 18.7 J mol^-1 K^-1.

(i) The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space.
(ii) Gas molecules are in constant random motion. Just as many molecule are moving in one direction as in any other.
(iii) Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas.
(iv) When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be perfectly elastic. The total kinetic energy of all the molecules remains constant unless there is some outside interference with the molecules.
(v) The molecule exert no attractive or repulsive forces on one another except during the process of collision. Between collisions, they move in straight lines.