Question

An ideal gas has molar heat capacity $C$, at constant volume. The molar heat capacity of this gas in the process $T=T_0{e}^{\alpha V}$ is

• A. $C_v+\frac{R}{{\alpha V}^2}$
• B. $C_v-\frac{R}{{\alpha V}^2}$
• C. $C_v+\frac{R}{\alpha V}$
• D. $C_v-\frac{R}{\alpha V}$

Answer 1

Answer: (C)

$C_v+\frac{R}{\alpha V}$

$\text{given}:T=To{e}^{\alpha }V$ $----\to A$

$\text{solution}$: we know that

$C\Delta T=Cv\Delta T+P\Delta V$ $----\to B$

(using 1 st law of thermodynamics )

1. $dQ=dU+dW$

2. $dU=CvdT$

3. $dW=PdV$ putting (2.) and (3. )in (1 .)we get $C\Delta T=Cv\Delta T+P\Delta V$

differentiating equation A from both side we get

$\frac{dT}{dV}=\alpha To{e}^{\alpha }V$

$\frac{dT}{dV}=\alpha T$ { $\therefore T=To{e}^{\alpha }V$}

eq. B can also be written as

$C=Cv+P\frac{dV}{dT}$

this can also be written as

$C=Cv+P\frac{1}{\alpha T}$ $----\to C$

from ideal gas equation

$PV=RT$

$\frac{P}{T}=\frac{R}{V}$

by putting this value in equation C we get

$C=Cv+\alpha \frac{R}{V}$

Hence , option c is correct