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(A) Let the initial pressure be $p$ and the initial number of moles be $n_1$. According to the ideal gas law,$pV = n_1RT$.After isothermal compression

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$66$

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(A) Let the initial pressure be $p$ and the initial number of moles be $n_1$. According to the ideal gas law,$pV = n_1RT$.After isothermal compression to volume $V/3$,the new pressure $p'$ becomes $p' = n_1RT / (V/3) = 3(n_1RT/V) = 3p$.Now,the valve is opened and gas leaks out until the pressure returns to the original value $p$,while the volume remains $V/3$ and temperature $T$ remains constant.Let the new number of moles be $n_2$. Then,$p(V/3) = n_2RT$.Dividing the two equations: $(pV) / (pV/3) = n_1 / n_2$,which gives $3 = n_1 / n_2$,or $n_2 = n_1 / 3$.The number of moles that escaped is $\Delta n = n_1 - n_2 = n_1 - n_1/3 = 2n_1/3$.The percentage of molecules that escaped is $(\Delta n / n_1) 	imes 100 = (2/3) 	imes 100 \approx 66 \%$.

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