Question

The molar specific heat of a gas as given from the kinetic theory is $\dfrac{{5R}}{2}$. If it is not specified whether it is ${C_P}$ or ${C_V}$​, one could conclude that the molecules of the gas.
A. Are definitely monoatomic
B. Are definitely rigid diatomic
C. Are definitely non-rigid diatomic
D. Can be monatomic or rigid diatomic

Answer 1

Hint: When there is a single gaseous atom it can move in all 3 directions. That means along x axis and along y axis and along z axis. So for the single atom the degrees of freedom will be three. Similarly we can determine the degrees of freedom for diatomic, triatomic and polyatomic molecules too. Based on the degrees of freedom only we will answer the question.
Formula used:
$\eqalign{
  & {C_V} = \dfrac{{fR}}{2} \cr
  & {C_P} = {C_V} + R \cr} $

Complete answer:
So the monoatomic molecule can move in all three directions. So there will be translational kinetic energy in all three directions. Hence degrees of freedom of monoatomic molecules is 3. If we consider a diatomic molecule i.e rigid diatomic molecule there are three translational energy modes and only two rotational energy modes because since diatomic molecule will be linear, it can translate in all three directions but the rotational kinetic energy will be only considered along any two axes perpendicular to the linear alignment of the atoms of the linear molecule. So a total of five degrees of freedom. If we consider the non rigid diatomic molecule then there will be one additional degree of freedom i.e vibrational, because the internuclear distance between the two non rigid atoms can be varied due to vibrations. So total there will be 6 degrees of freedom for non rigid diatomic molecules.
So in conclusion the molar specific heat at constant volume (${C_V}$) is given as
${C_V} = \dfrac{{fR}}{2}$ where ‘f’ is the degrees of freedom and ‘R’ is the universal gas constant.
${C_V} = \dfrac{{fR}}{2}$
$\therefore {C_V} = \dfrac{{3R}}{2}$
We have another relation which relates molar specific heat at constant volume (${C_V}$) and molar specific heat at constant pressure (${C_P}$) which is given by ${C_P} = {C_V} + R$
$\eqalign{
  & {C_P} = {C_V} + R \cr
  & \Rightarrow {C_P} = \dfrac{{3R}}{2} + R \cr
  & \therefore {C_P} = \dfrac{{5R}}{2} \cr} $
For rigid diatomic we have
${C_V} = \dfrac{{fR}}{2}$
$\therefore {C_V} = \dfrac{{5R}}{2}$
So it can be monatomic or rigid diatomic.

Hence option D will be the correct one.

Note:
In the question it is clearly mentioned as gaseous molecules because gaseous molecules can freely translate and rotate and vibrate. If we consider solid molecules then we know that atoms of solids are closely packed and there will be no space for translation and rotation. There will be only vibration of solid atoms hence only vibrational degrees of freedom.