Question

Consider the following $PV$ diagram. $\text{AB}$ is an isothermal process during which the temperature is $400\text{K}$. $\text{BC}$ is an isochoric process and $\text{CA}$ is an adiabatic process. If $1\text{mole}$ of diatomic gas undergoes the cycle, then find the total work done by the gas. $[\text{ln}2=0.7]$

• A. $-110R$
• B. $220R$
• C. $110R$
• D. $-220R$

Answer 1

The correct option is D $-220R$

For the process $\text{AB}$ (isothermal process),

$W=nRT\text{ln}\frac{V_2}{V_1}$

$\Rightarrow W_1=1\times R\times 400\times \ln \left(\frac{2V}{V}\right)$

$\Rightarrow W_1=400R\ln 2=280R.....\left(1\right)$

For the process $\text{BC}$ (isochoric process),

$\Delta V=0\Rightarrow W_2=0.....\left(2\right)$

For the process $\text{CA}$ (adiabatic process),

$W_3=\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }$

$\Rightarrow W_3=\frac{P\times V-\frac{P}{4}\times 2V}{1-\gamma }=\frac{PV}{2(1-\gamma )}$

$\Rightarrow W_3=\frac{RT}{2(1-\gamma )}[PV=nRT,n=1]$

$\Rightarrow W_3=\frac{R\times 400}{2(1-1.4)}$

[For diatomic gas, $\gamma =1.4]$

$\Rightarrow W_3=-500R.....\left(3\right)$


So, total work done in the process,

$W=280R+0+(-500R)$ [From $\left(1\right)$ , $\left(2\right)$ and $\left(3\right)$]

$\Rightarrow W=-220\text{R}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip - For a cyclic process, first find the work done in all}\\ \text{the individual processes and then add them to get}\\ \text{the total work done.}\\ \text{Caution - If PV graph is simple, then just find the area}\\ \text{under the curve to get the work done.}\end{array}$