JEE MAIN - Physics (2024 - 6th April Morning Shift - No. 2)
The specific heat at constant pressure of a real gas obeying $$P V^2=R T$$ equation is:
R
$$C_V+R$$
$$C_V+\frac{R}{2 V}$$
$$\frac{R}{3}+C_V$$
Explanation
$$\begin{aligned} & \because \quad P V^2=R T \\ & P(2 v d v)+V^2(d P)=R d T \end{aligned}$$
at $$P=$$ const.
$$P d v=\frac{R d T}{2 V} \quad \text{... (i)}$$
Now, for $$n=1$$
$$\begin{aligned} & d \theta=d v+d w \\ & C_P d T=C_v d T+P d v \quad \text{... (ii)} \end{aligned}$$
from (i) and (ii)
$$C_P=C_V+\frac{R}{2 V}$$
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