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.
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