Question

An ideal gas is taken through the cycle $ABCA$, as shown in the figure. What will be the work done by the gas in process CA (in J) if the net heat supplied to the gas in the cycle is $15J$ ?

Answer 1

Answer: -15

For cyclic process,
Total work done $=W_{AB}+W_{BC}+W_{CA}$
$\therefore \Delta W_{AB}=P\Delta V=20(3-1.5)=30J$
and $\Delta W_{BC}=0$
From $1^{\text{st }}$ law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
$\Delta U=0$ (Process $ABCA$ is cyclic)
$\therefore \Delta Q=\Delta W_{AB}+\Delta W_{BC}+\Delta W_{CA}$
$\therefore 15=30+0+\Delta W_{CA}$
$\therefore \Delta W_{CA}=-15J$