Question

Relationship between the ratio of specific heats ( $\gamma$ ) of gas and degree of freedom $'f'$ will be
A. $\gamma = f + 2$
B. $\dfrac{1}{\gamma } = \dfrac{1}{f} + \dfrac{1}{2}$
C. $f = \dfrac{2}{{\gamma - 1}}$
D. $f = 2\left( {\gamma - 1} \right)$

Answer 1

In order to solve this you have to remember the concepts of the specific heats and the degree of freedom and also the relationship between the ratio of specific heats of gas and the degree of freedom. And also recall the formula for specific heat at constant pressure and constant volume and their relationship.

Formula used:
The ratio of specific heats ( $\gamma$ ) of gas is given by
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$
Where, ${C_P}$​ is the specific heat at constant pressure
${C_V}$ is the specific heat at constant volume
The internal energy with degree of freedom $'f'$, is given by
$U = \dfrac{f}{2}{K_B}T$
Where, $U$ is the internal energy
$f$ is the degree of freedom
${K_B}$​ is Boltzmann's constant
$T$ is the temperature

Complete step by step solution:
We know that the ratio of specific heats ( $\gamma$ ) of gas is given by
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$ ..............(i)
and the formula for internal energy with degree of freedom $'f'$, is given by
$U = \dfrac{f}{2}{K_B}T$ …………….(ii)
We know that the total energy is
$E = U \times N$ ……………..(iii)
Where, $N$ is the no. of molecules
Now putting the value from equation (ii)
$E = \dfrac{f}{2}N{K_B}T$
Now take $R = N{K_B}$, we have
$\Rightarrow E = \dfrac{f}{2}RT$ ………….(iv)
Now, we know that the specific heat at constant volume is given by,
${C_V} = {\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_V}$
On putting value of E from equation (iv) and then further solve, we have
$\Rightarrow {C_V} = \dfrac{f}{2}R$ …………..(v)
Now the relation between the specific heats is given by,
${C_P} - {C_V} = R$
On putting the value of ${C_V}$ in the above equation from equation (v), we get
${C_P} = \dfrac{R}{2}\left( {f + 2} \right)$ …………….(vi)
Now put the values from equation (v) and (vi) in equation (i),
$\gamma = \dfrac{{\dfrac{R}{2}\left( {f + 2} \right)}}{{\dfrac{R}{2}fT}}$
$\Rightarrow \gamma = 1 + \dfrac{2}{f}$
On further solving we have,
$f = \dfrac{2}{{\gamma - 1}}$
Hence, this is the relationship between the ratio of specific heats ( $\gamma$ ) of gas and degree of freedom $'f'$.

Therefore, the correct option is C

Note: The specific heat at constant pressure is greater than the specific heat at the constant volume, because when the heat is added at constant pressure, the material expands and works whereas when we add a gas at constant volume then no work is done.