Chemistry - Thermodynamics - 1 Question with Solution | TestHub

We know that,
$\mathrm{\Delta }\mathrm{u}=0$in cyclic process
So$\mathrm{\Delta }\mathrm{Q}=\mathrm{\Delta }\mathrm{u}+\mathrm{\Delta }\mathrm{w}$
$\mathrm{\Delta }\mathrm{Q}=\mathrm{\Delta }\mathrm{w}$
We know,
work done = Area under P-V curve
= Area of circle
$=\frac{\mathrm{\pi }{\mathrm{r}}^2}{4}$
$=\frac{\mathrm{\pi }}{4}\times (30-10)\times 10^3(30\mathrm{}-10)\times 10^{-3}$
$=\frac{\mathrm{\pi }}{4}\times 20\times 20$
$=\frac{\mathrm{\pi }}{4}\times 400$
$=100\mathrm{\pi }$
$=10^2\mathrm{\pi }\text{J}$
So Heat energy absorbed
$\mathrm{\Delta }\mathrm{Q}=\mathrm{\Delta }\mathrm{w}$
$=10^2\mathrm{\pi }\mathrm{}\mathrm{J}$