Question

The equation of a process of diatomic gas $P^2={\alpha }^{2}V$ where $\alpha$ is a positive constant. Then choose the incorrect option(s)

• A. Work done by gas for a temperature change $T$ is $\frac{2}{3}\alpha nRT$.
• B. The change in internal energy is $\frac{5}{2}nRT$ for a temperature change $T$.
• C. The specific heat for the process is $\frac{19}{9}R$.
• D. The change in internal energy for a temperature change $T$ is $\frac{5}{2}\alpha nRT$.

Answer 1

Answer: (B), (C)

The correct options are
B The change in internal energy is $\frac{5}{2}nRT$ for a temperature change $T$.
C The specific heat for the process is $\frac{19}{9}R$.

$P^2={\alpha }^{2}V$
$\Rightarrow P=\alpha \sqrt{V}$
$W=\underset{}{\overset{}{\int }}PdV=\underset{V_1}{\overset{V_2}{\int }}\alpha \sqrt{V}dV$
$=\alpha \frac{2}{3}(V_2^{3/2}-V_1^{3/2})$ ---(1)

$PV=nRT$
$\Rightarrow \alpha \sqrt{V}V=nRT$
$\Rightarrow {\left(\frac{V_2}{V_1}\right)}^{3/2}=\frac{T_2}{T_1}$ --- (2)
$\Rightarrow \frac{(V_2{)}^{3/2}-(V_1{)}^{3/2}}{V_1^{3/2}}=\frac{T_2-T_1}{T_1}=\frac{T}{T_1}$---(3)

From (1) , (2) and (3) ,

$W=\frac{2}{3}\alpha \frac{T}{T_1}{V}_1^{3/2}=\frac{2}{3}\alpha T\frac{nR}{\alpha }$ $[\alpha V_1^{3/2}=nRT,\frac{V_1^{3/2}}{T_1}=\frac{nR}{\alpha }]$
$=\frac{2}{3}nRT$

Alternate:
When $P{V}^x=$constant , $W=\frac{nRdT}{1-x}$
Given $P{V}^{-1/2}=\alpha \Rightarrow W=\frac{nRT}{1+\frac{1}{2}}=\frac{2}{3}nRT$

Change in internal energy for change in temperature $T$ $dU=n{C}_{V}dT=\frac{n5R}{2}T=\frac{5}{2}nRT$
Option B is correct.

$Q=U+W$
$\Rightarrow \frac{dQ}{dT}=\frac{dU}{dT}+\frac{dW}{dT}$

Thus, specific heat for the process
$C=C_V+\frac{R}{1-x}=\frac{5}{2}R+\frac{2}{3}R=\frac{19}{6}R$
Option C is correct.