Question

The translational kinetic energy of molecules of one mole of a monatomic gas is U= 3NkT/2. The value of atomic specific heat of gas under constant pressure will be :
A) \[\dfrac{3}{2}R\]
B) \[\dfrac{5}{2}R\]
C) \[\dfrac{7}{2}R\]
D) \[\dfrac{9}{2}R\]

Answer 1

Hint:From Maxwell’s law of equipartition of energy, kinetic energy associated with each degree of freedom of particles of an ideal gas is equal to \[\dfrac{1}{2}kT\]. For monoatomic gas ( He, Ar etc), the degree of freedom i.e. \[f\]=3.

Formula used:- The value of atomic specific heat of gas under constant pressure is \[{C_P} = {C_V} + R\], also the value of atomic specific heat of gas under constant volume \[{C_V}\]is, by mathematical relation, \[{C_V} = \dfrac{f}{2}R\].
Degree of freedom (\[f\]) :- It is the minimum coordinates required to specify the dynamical state of a system.
For monoatomic gas ( \[{C_V}\]He, Ar etc.) \[f\]=3 , as they have only translational degrees of freedom.
For diatomic gas (\[{H_{2,}}\], etc) \[f\]=5 , as they have 3 translational degrees of freedom and 2 rotational degrees of freedom .
Average kinetic energy ( K.E ) of a particle having \[f\] degree of freedom = \[\dfrac{f}{2}kT\]
Translational kinetic energy( K.E ) of a molecule =\[\dfrac{3}{2}kT\]

Step by step solution :-
The value of atomic specific heat of gas under constant pressure is \[{C_P} = {C_V} + R\]
also value of atomic specific heat of gas under constant volume is
by mathematical relation \[{C_V} = \dfrac{f}{2}R\].
For monoatomic gases , \[f\]=3 , as they have only translational degrees of freedom.
\[{C_V} = \dfrac{f}{2}R = \dfrac{3}{2}R\]
The value of atomic specific heat of gas under constant pressure is \[{C_P} = {C_V} + R\]
\[{C_P} = {C_V} + R = \dfrac{3}{2}R + R = \dfrac{5}{2}R\]
Since , The value of ato\[ \Rightarrow N = 1\]mic specific heat of gas under constant pressure is \[{C_P} = \dfrac{5}{2}R\]

Option ( B ) is the correct answer.

Note:- Kinetic interpretation of temperature :
Temperature of an ideal gas is proportional to the average K.E of molecules.
\[PV = \dfrac{1}{3}mN{V_{rms}}^2\] &
If we are making the relation for one mole of gas

\[PV = \dfrac{1}{3}m{V_{rms}}^2 = \dfrac{3}{2}kT\].