Question

An ideal gas undergoes a cyclic process as shown in Figure.
$\Delta U_{BC} = - 5 \, kJ \, mol^{-1} , q_{AB} = 2 \, kJ \, mol^{-1}$
$W_{AB} = - 5 \, kJ \, mol^{-1} , W_{CA} = 3 \, kJ \, mol^{-1}$
Heat absorbed by the system during process $CA$ is :

• A. $-5 \, kJ \, mol^{-1}$
• B. $ + 5 \, kJ \, mol^{-1}$
• C. $18 \, kJ \, mol^{-1}$
• D. $ - 18 \, kJ \, mol^{-1}$

Answer 1

Answer: (B)

We have $\Delta U_{ AB } =q_{ AB }+w_{ AB }$

$=2+(-5)=-3\, kJ\, mol ^{-1}$

For a cyclic process, $\Delta U=0 .$

Therefore, $\Delta U=\Delta U_{A B}+\Delta U_{B C}+\Delta U_{C A}$

$\Delta U_{ CA } =-\Delta U_{ AB }-\Delta U_{ BC }$

$=-(-3)-(-5)=8\, kJ\, mol ^{-1}$

$\Delta U_{ CA } =q_{ CA }+w_{ CA }$

$8 =q_{ CA }+3 \Rightarrow q_{ CA }=5\, kJ\, mol ^{-1}$