Question

A cyclic process for $1$ mole of an ideal gas is shown in figure in the $V-T$ diagram. The work done in process $AB,BC$ and $CA$ respectively, are

• A. $0,2R{T}_2\ln \left(\frac{V_1}{V_2}\right),R(T_1-T_2)$
• B. $R(T_1-T_2),0,R{T}_1\ln \frac{V_1}{V_2}$
• C. $0,R{T}_2\ln \left(\frac{V_2}{V_1}\right),-2R(T_1-T_2)$
• D. $0,R{T}_2\ln \left(\frac{V_2}{V_1}\right),-R(T_2-T_1)$

Answer 1

The correct option is D $0,R{T}_2\ln \left(\frac{V_2}{V_1}\right),-R(T_2-T_1)$

From the $V-T$ graph,

In process $AB$, volume is constant.
As volume is constant, work done in this process is $0$.

In process $BC$, temperature is constant.
Work done in isothermal process is
$W_{BC}=R{T}_2\ln \left(\frac{V_2}{V_1}\right)$

In process $CA$, $V\propto T$
or $\frac{V}{T}$ is constant i.e pressure is constant.
[from $PV=nRT$]
$\therefore W_{CA}=P\Delta V=R(T_1-T_2)=-R(T_2-T_1)$
Thus, option (d) has the correct graph.