Question

Relation between the ratio of specific heats ($\gamma$) of gas and degree of freedom 'f' will be

• A. $\gamma =f+2$
• B. $\frac{1}{\gamma }=\frac{1}{f}+\frac{1}{2}$
• C. $f=2/(\gamma -1)$
• D. $f=2(\gamma -1)$

Answer 1

The correct option is C $f=2/(\gamma -1)$

$\gamma =\frac{C_P}{C_V}$ where,

$C_P$ is the specific heat at constant pressure.

$C_V$ is the specific heat at constant volume.

$U=\frac{f}{2}{K}_{B}T$ where,

$U$ is the internal energy

$f$ is the degree of freedom

$K_B$ is Boltzmann's constant

$T$ is temperature.

$E=U\times N$ (N=no. of molecules, $R=N{K}_B$)

$=\frac{f}{2}RT$

Now, $C_V={\left(\frac{\partial E}{\partial T}\right)}_V=\frac{f}{2}R$

$C_P-C_V=R$

$C_P=C_V+R$

$=\frac{R}{2}(f+2)$

$\gamma =\frac{C_P}{C_V}=\frac{f+2}{f}=1+\frac{2}{f}$

$\gamma f=f+2$

$f=\frac{2}{\gamma -1}$