Two moles of an ideal gas undergo free expans | vedclass

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(D) For an ideal gas undergoing free expansion, the external pressure $P_{ext} = 0$. Since $w = -P_{ext} \Delta V$, the work done $w = 0$. According t

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$\Delta S_{system} = 4.606R; \Delta S_{surroundings} = 0$

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(D) For an ideal gas undergoing free expansion, the external pressure $P_{ext} = 0$. Since $w = -P_{ext} \Delta V$, the work done $w = 0$. According to the first law of thermodynamics, $\Delta U = q + w$. For an ideal gas, $\Delta U = 0$ at constant temperature, so $q = 0$. Since $q = 0$, the process is adiabatic. For the surroundings, $\Delta S_{surr} = -q_{rev}/T = 0$ because $q = 0$. For the system, entropy change is given by $\Delta S_{sys} = nR \ln(V_2/V_1)$. Given $n = 2$, $V_1 = 10 \ L$, $V_2 = 100 \ L$. $\Delta S_{sys} = 2 	imes R 	imes \ln(100/10) = 2R \ln(10) = 2 	imes 2.303 	imes R 	imes \log_{10}(10) = 4.606R$. Thus, $\Delta S_{system} = 4.606R$ and $\Delta S_{surroundings} = 0$.

Two moles of an ideal gas undergo free expansion from $10 \ L$ to $100 \ L$ at $300 \ K$. The values of $\Delta S_{system}$ and $\Delta S_{surroundings}$ are ($R$ is universal gas constant)

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