A gas in a closed container undergoes the cycle $\mathrm{ABCA}$ as shown in the figure. The net heat…
A gas in a closed container undergoes the cycle $\mathrm{ABCA}$ as shown in the figure. The net heat absorbed by the gas after it has completed 10 cycles is
$-1.5 \mathrm{~kJ}$
$+1.5 \mathrm{~kJ}$
$+2.25 \mathrm{~kJ}$
$-2.25 \mathrm{~kJ}$
Solution
$\Delta \mathrm{Q}=\Delta \mathrm{W} \quad[\because$ process is cyclic $]$
$=-\left[\frac{1}{2} \times 15 \times 20\right] \times 10 \quad[\because$ no. of cycles $=10]$
$=-1500 \mathrm{~J}$
As cycle is anti-clock wise. So $\Delta \mathrm{W}$ is -ve therefore, $\Delta \mathrm{Q}$ is -ve.
