EK, Chemistry, In-Class exam 3 #69

[somefun]

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Two ideal gases, A and B, are at the same temperature, volume, and pressure. Gas A is reversibly expanded at constant temperature to a volume V. Gas B is allowed to expand into an evacuated chamber until it also has a total volume V, but without exchanging heat with its surroundings. Which of the following most accurately describes the two gases?

A) Gas A has a higher temperature and enthalpy than gas B.
B) Gas A has a higher temperature but a lower enthalpy than gas B.
C) Gas B has a higher temperature and enthalpy than gas A.
D) Gas A and B have equal temperature and enthalpies



Answer is D)

EK explanation: D is correct. The temperature of Gas A remains constant because the question says so. Temperature is kinetic energy per mole. Gas B does no work and doesn't exchange heat so its energy doesn't change; it has the same kinetic energy per mol as when it began. Thus, its temperature doesn't change either.
Enthalpy is PV+U. P and V are the same for both gases because they are at the same temp, volume, and therefore pressure (PV = nRT). U doesn't change for Gas A because any energy removed is replaced to keep the temperature the same. U doesn't change for Gas B because no energy is exchanged with the surroundings for Gas B.

My question:
I understand that Gas A will retain its temperature. But I don't get how Gas B can expand into a bigger volume without changing its pressure OR volume OR temperature. It's stated that it never interacts with the outside environment, so how could something increase in volume without a change in other properties?

This is what how I understood this concept, but it took me a while so I would like someone else's input and correction if needed.

The Pressure and Volume do change, the explanation is saying that the pressure and volume is the same in comparison to both gases. However, TEMPERATURE doesn't change.
In order for temperature to change energy would have to be changed as well. How can you change/transfer energy? Through heat or work. Since work is not done (because although it is expanded it is expanded in a evacuated system - meaning that there is no force against the expansion, hence no work) nor heat is exchanged with the surroundings, energy must remain the same indicating that temperature also stays the same.

 

[somefun]

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[Gauss44]

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[Czarcasm]

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Gas A: constant temperature. Since temperature is proportional to the average kinetic energy (internal energy of the system), this means deltaE is zero and -w=q (energy lost to expand the container is gained back as heat added to the system from surroundings). Gas B describes an adiabatic process (q=0), and because there is no work done by the gas to expand the container, the internal energy also remains constant. In both scenarios, because internal energy is constant, temperature should remain the same.

Not entirely sure if this is a valid approach, but that's how I arrived at D.