Question

$n$ moles of a perfect gas undergoes a cyclic process$ABCA$ (see figure) consisting of the following processes:

$A\to B$: Isothermal expansion at a temperature$T$ so that the volume is doubled$V_{1}to{V}_2$ and pressure changes$P_{1}to{P}_2$.

$B\to C$: Isobaric compression at pressure$P_2$ to initial volume$V_1$.

$C\to A$: Isochoric change leading to change of pressure from$P_{1}to{P}_2$.

Total work done in the complete cycle$ABCA$ is –

• A. $0$
• B. $nRT(\ln 2+\frac{1}{2})$
• C. $nRT\ln 2$
• D. $nRT(\ln 2–\frac{1}{2})$

Answer 1

The correct option is D

$nRT(\ln 2–\frac{1}{2})$

Step 1. Given data:

1. $A\to B$: Isothermal expansion at a temperature$T$, volume is doubled$V_{1}to{V}_2$, and pressure changes$P_{1}to{P}_2$.
2. $B\to C$: Isobaric compression at pressure$P_2$ to initial volume$V_1$.
3. $C\to A$: Isochoric change leading to change of pressure from$P_{1}to{P}_2$.

Step 2. Calculating work done in the isothermal process:

At a constant temperature$T$, the work done is,

$$\begin{gathered} dW=PdV \\ W={\int }_{V_1}^{V_2}PdV \end{gathered}$$ (where, $dV=$The change in volume)

From ideal gas equations for$n$ moles of a perfect gas,

$\begin{array}{rcl}PV& =& nRT\\ & \Rightarrow & P=\frac{nRT}{V}\\ W& =& {\int }_{V_1}^{V_2}\frac{nRT}{V}dV\\ & =& nRT{\int }_{V_1}^{V_2}\frac{dV}{V}\\ & =& nRT\ln \frac{V_2}{V_1}\end{array}$ (where $R=$Universal gas constant)

$\begin{array}{rcl}W& =& nRT\ln \frac{2V_1}{V_1}\\ & =& nRT\ln 2(V_2=2V_1)\end{array}$

Step 3. Calculating work done in the isobaric process:

In the isobaric process under constant pressure, the work done is,

$\begin{array}{rcl}dW& =& P_{2}dV=nRdT\\ W& =& {\int }_T^{\frac{T}{2}}nRdT(\therefore V\to \frac{V}{2})\\ & =& nR\left(\frac{T}{2}-T\right)\\ & =& -\frac{nRT}{2}\\ & & \end{array}$

Step 4. Calculating work done in the isochoric process:

In the isochoric process under constant volume, the work done is,

$\begin{array}{rcl}W& =& PdV\\ & =& 0(dV=0)\end{array}$

Step 4. Calculating total work done:

The total work done we can get by adding the work done during each step.

$\begin{array}{rcl}W& =& nRT\ln 2-\frac{nRT}{2}+0\\ & =& nRT\left(\ln 2-\frac{1}{2}\right)\end{array}$