Question

$n$ moles of an ideal diatomic gas undergoes a process in which the pressure of the gas is proportional to its volume. If the temperature of gas changes by $\Delta T$, choose the correct statements.

• A. The work done will be, $\frac{nR\left(\Delta T\right)}{2}$
• B. The internal energy change for gas will be, $\frac{3nR\left(\Delta T\right)}{2}$
• C. The molar specific heat capacity for gas will be $3R$
• D. Given data is insufficient

Answer 1

Answer: (A), (C)

The correct options are
A The work done will be, $\frac{nR\left(\Delta T\right)}{2}$
C The molar specific heat capacity for gas will be $3R$

Work done, $\Delta W={\int }_{V_1}^{V_2}PdV$
Given:
Pressure of the gas $\propto$ volume
$\Rightarrow P=kV$ where $k$ is a constant]

$\therefore \Delta W={\int }_{V_1}^{V_2}kVdV=\frac{k}{2}[V_2^2-V_1^2]$

$\Delta W=\frac{\left(K{V}_2\right)\left(V_2\right)-\left(K{V}_1\right)\left(V_1\right)}{2}=\frac{P_2{V}_2-P_1{V}_1}{2}$

$\Delta W=\frac{nR{T}_2-nR{T}_1}{2}=\frac{nR\Delta T}{2}$
For diatomic gas $C_v=\frac{5R}{2}$
Internal energy change of gas is,
$\Delta U=n{C}_v\Delta T=n\left(\frac{5R}{2}\right)\Delta T$
First law of thermodynamics gives us,
$\Rightarrow \Delta Q=\Delta U+\Delta W$
$\Rightarrow nC\Delta T=n\left(\frac{5}{2}R\right)\Delta T+nR\frac{\Delta T}{2}$
$\Rightarrow C=3R$