A cylinder of ideal gas is closed by an $8.00$ kg movable piston $\left(\right.$ area $\left.=60.0 \mathrm{~cm}^{2}\right)$ as illustrated in $\underline{\text { Fig. } 20-5}$. Atmospheric pressure is $100 \mathrm{kPa}$. When the gas is heated from $30.0^{\circ} \mathrm{C}$ to $100.0^{\circ} \mathrm{C}$, the piston rises $20.0 \mathrm{~cm}$. The piston is then fastened in place, and the gas is cooled back to $30.0^{\circ} \mathrm{C}$. Calling $\Delta Q_{1}$ the heat added to the gas in the heating process, and $\Delta Q_{2}$ the heat lost during cooling, find the difference between $\Delta Q_{1}$ and $\Delta Q_{2}$.
During the heating process, the internal energy changed by $\Delta U_{1}$, and an amount of work $\Delta W_{1}$ was done. The absolute pressure of
the gas was
$$P=\frac{m g}{A}+P_{A}$$
$$P=\frac{(8.00)(9.81) \mathrm{N}}{60.0 \times 10^{-4} \mathrm{~m}^{2}}+1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.13 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$$
Therefore,
$$\begin{aligned}
\Delta Q_{1} &=\Delta U_{1}+\Delta W_{1}=\Delta U_{1}+P \Delta V \\
&=\Delta U_{1}+\left(1.13 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.200 \times 60.0 \times 10^{-4} \mathrm{~m}^{3}\right)=\Delta U_{1}+136 \mathrm{~J}
\end{aligned}$$
During the cooling process, $\Delta W=0$ and so (since $\Delta Q_{2}$ is heat lost)
$$-\Delta Q_{2}=\Delta U_{2}$$
But the ideal gas returns to its original temperature, and so its internal energy is the same as at the start. Therefore $\Delta U_{2}=-\Delta U_{1}$, or $\Delta Q_{2}=\Delta U_{1} .$ It follows that $\Delta Q_{1}$ exceeds $\Delta Q_{2}$ by $136 \mathrm{~J}=32.5$ cal.