Question

A certain quantity of an ideal gas taken up $56J$ of heat in the process $AB$ and $360J$in the process $AC$, as shown in the figure. What is the number of degrees of freedom of the gas?

Answer 1

Given,
$Q_{AB}=56J$
$Q_{AC}=360J$
For process $AB$,
$Q_{AB}=n{C}_P\Delta T=\frac{\gamma }{\gamma -1}nR\Delta T$
$\Rightarrow Q_{AB}=\frac{\gamma }{\gamma -1}\left[2p_0{V}_0\right]=2P_0{V}_0\times \frac{\gamma }{\gamma -1}$
$\Rightarrow 56=2P_0{V}_0\times \frac{\gamma }{\gamma -1}..\left(i\right)$
For process $AC$,
$Q_{AC}=\Delta U+W$
$Q_{AC}=\frac{nR}{\gamma -1}\Delta T+\frac{1}{2}\times 3V_0[P_0+4P_0]$
$Q_{AC}=\frac{[16P_0{V}_0-P_0{V}_0]}{\gamma -1}+\frac{15P_0{V}_0}{2}$
$\Rightarrow 360=15P_0{V}_0\left[\frac{\gamma +1}{2(\gamma -1)}\right]\dot{.}..\left(ii\right)$
(𝑖𝑖)÷(𝑖), we get
$\frac{360}{56}=\frac{15}{4}\frac{(\gamma +1)}{\gamma }$
$12\gamma =7\gamma +7$
$\gamma =\frac{7}{5}=1+\frac{2}{f}$
$\Rightarrow f=5$
Final answer: (5)