Question

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

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

Answer 1

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

From indicator diagram ($V-T$ graph)
Process $AB$ is isochoric.
$\therefore W_{AB}=0$
Process $BC$ is isothermal,
$\therefore W_{BC}=R{T}_2\ln \left(\frac{V_2}{V_1}\right)$
Process $CA$ is isobaric,
$\therefore W_{CA}=P(V_1-V_2)$
Using, $PV=RT$
$W_{CA}=R(T_1-T_2)$