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(B) When an ideal gas expands into a vacuum,the external pressure $P_{ext} = 0$. Since work done $w = -P_{ext} \Delta V$,the work done is $0$. For an

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$\Delta S > 0$

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(B) When an ideal gas expands into a vacuum,the external pressure $P_{ext} = 0$. Since work done $w = -P_{ext} \Delta V$,the work done is $0$. For an adiabatic process in a vacuum,$q = 0$. According to the first law of thermodynamics,$\Delta U = q + w = 0$. For an ideal gas,$\Delta U$ depends only on temperature,so $\Delta T = 0$. However,the expansion into a vacuum is an irreversible process. The entropy change for an ideal gas is given by $\Delta S = nR \ln \frac{V_2}{V_1}$. Since $V_2 > V_1$,$\Delta S > 0$.

What is the change in entropy when an ideal gas expands into a vacuum?

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