Question

The diagram shows a cyclic process performed on one mole of an ideal gas. A total of $1000J$ of heat is withdrawn from the gas in a complete cycle. Find the work done by the gas in the process $B\to C$

• A. $831J$
• B. $1831J$
• C. $-1831J$
• D. $-1000J$

Answer 1

The correct option is C $-1831J$


Given, $T_1=300K,T_2=400K,\Delta Q=-1000J$ and $\Delta U=0$

Using first law of thermodynamics, $\Delta Q=\Delta U+\Delta W$
Here, $\Delta U=0$,
$\Delta Q=\Delta W\Delta Q=W_{AB}+W_{BC}+W_{CA}$

Process $CA$ is a isochoric process because volume is constant, hence $W_{CA}=0$

$\Delta Q=W_{AB}+W_{BC}$

and process $AB$ is an isobaric process. So $P\propto V$
$\Delta Q=P(V_2-V_1)+W_{BC}$

Using ideal gas equation: $PV=nRT$

$-1000=nR(T_2-T_1)+W_{BC}-1000=R(400-300)+W_{BC}(n=1)-1000=8.31\times 100+W_{BC}$
$\therefore W_{BC}=-1831J$