Question

A cyclic process for one mole of an ideal gas is shown in the $V-T$ diagram. The work done in $AB,BC$ and $CA$ respectively are

• A. $0,R{T}_{1}ln\left(\frac{V_1}{V_2}\right),R(T_1-T_2)$
• B. $R,(T_1-T_2)R,R{T}_{1}ln\left(\frac{V_1}{V_2}\right)$
• C. $0,R{T}_{1}ln\left(\frac{V_2}{V_1}\right),\frac{R{T}_1}{V_1}(V_1-V_2)$
• D. $0,R{T}_{2}ln\left(\frac{V_2}{V_1}\right),R(T_2-T_1)$

Answer 1

The correct option is C $0,R{T}_{1}ln\left(\frac{V_2}{V_1}\right),\frac{R{T}_1}{V_1}(V_1-V_2)$

$\text{Explanation:-}$

During AB, process is isochoric

$\therefore \Delta V=0$ $\therefore W=0$

During BC, process is isothermal

$\therefore \Delta T=0$ $\therefore W=R{T}_{2}ln\frac{V_2}{V_1}$

During CA, process is isobaric. So, pressure is constant,

$\therefore W=P(V_1-V_2)$

But $P{V}_1=R{T}_1$

$\therefore P=\frac{R{T}_1}{V_1}=\frac{R{T}_2}{V_2}$

$\therefore W=\frac{R{T}_1}{V_1}(V_1-V_2)=\frac{R{T}_2}{V_2}(V_1-V_2)$

$\text{Hence the correct option is C}$