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

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

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