Question

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

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

Answer 1

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


$W_{AB}=0$ (Constant volume process.)

$W_{BC}\to isothermal=RT\ln \frac{V_2}{V_1}$

$W_{CA}\to isobaric=-P(V_C-V_A)=-R(T_2-T_1)$