A container of volume V is divided into two parts by a partition. One part contains n moles of an ideal gas at temperature T, and the other is vacuum. If the partition is removed, what is the change in entropy of the gas?

B. nR ( V_1+V_2V_1) — This is a free expansion process. Since the internal energy of an ideal gas depends only on temperature and the process is adiabatic, the temperature remains constant, and the entropy change is given by nR ln(V_final/V_initial).

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$A container of volume V is divided into two parts by a partition. One part contains n moles of an ideal gas at temperature T in volume V_{1}, and the other is vacuum with volume V_{2} . If the partition is removed, what is the change in entropy Δ S ?$

A. $n R ln (\frac{V _{1}}{V _{2}})$
B. $n R ln (\frac{V _{1} + V _{2}}{V _{1}})$
C. $n R ln (\frac{V _{2}}{V _{1}})$
D. $0$

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Correct answer: B — $n R ln (\frac{V _{1} + V _{2}}{V _{1}})$

Solution

For free expansion, Δ U = 0 ⟹ Δ T = 0 . Δ S = n R ln (\frac{V _{f}}{V _{i}}) = n R ln (\frac{V _{1} + V _{2}}{V _{1}}) .

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A container of volume V is divided into two parts by a partition. One part contains n moles of an ideal gas at temperature T in volume V1​, and the other is vacuum with volume V2​. If the partition is removed, what is the change in entropy ΔS?

Solution

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More Kinetic Theory of Gases practice questions

$A container of volume V is divided into two parts by a partition. One part contains n moles of an ideal gas at temperature T in volume V_{1}, and the other is vacuum with volume V_{2} . If the partition is removed, what is the change in entropy Δ S ?$