The gas pressure $=\frac{ \mathrm{Weightofpiston} }{ \mathrm{Areaofcross}-\mathrm{section} }+ \mathrm{atm}.\mathrm{pressure}$
$\begin{array}{l}=\frac{8\times 9.8}{60\times {10}^{-4}}+1.00\times {10}^5\mathrm{N}/{\mathrm{m}}^2\\ =1.13\times {10}^5\mathrm{N}/{\mathrm{m}}^2\end{array}$
During the heating process, the internal energy is changed by $∆{\mathrm{U}}_1$ and work $∆{\mathrm{W}}_1$ , is
done.
Therefore,${\mathrm{\Delta Q}}_1={\mathrm{\Delta U}}_1+{\mathrm{\Delta W}}_1={\mathrm{\Delta U}}_1+\mathrm{P}.\mathrm{dV}$
$\begin{array}{l}={\mathrm{\Delta U}}_1+\left(1.13+{10}^5\right)\left(0.20\times 60\times {10}^{-4}\right)\\ ={\mathrm{\Delta U}}_1+136\mathrm{J}\end{array}$
During the cooling process, no work is done as volume is constant, $\mathrm{\Delta W}=0$.
Hence, ${\mathrm{\Delta Q}}_2={\mathrm{\Delta U}}_2.\text{ But }{\mathrm{\Delta U}}_2$
But ${\text{ΔU}}_2$ is negative as the temperature decreases, and since the gas
returns to its original temperature, ${\mathrm{\Delta U}}_2=-{\mathrm{\Delta U}}_1$
Hence,$\left[{\mathrm{\Delta Q}}_1-\left|{\mathrm{\Delta Q}}_2\right|\right]$
$=\left({\mathrm{\Delta U}}_1+136-{\mathrm{\Delta U}}_1\right)=+136\mathrm{J}$