For the expansion of an ideal gas into a vacuum at constant temperature,ΔH = 0.What can be said about ΔE and ΔS?

Question

Quizplus · Accepted Answer

For an expansion of ideal gas into a vacuum, no work is done as there is no opposing pressure. Therefore, ΔE = q, and since ΔH = 0, q = 0 as well. Hence, ΔE = 0. As the entropy of the system increases due to the dispersion of gas particles in the vacuum, ΔS must be positive. Therefore, ΔE is zero, and ΔS is positive. Choice B is the only one that satisfies both conditions.

For the Expansion of an Ideal Gas into a Vacuum