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The question:

Enclosed beneath the movable piston in the drawing is 4.8 moles of a mono-atomic ideal gas. The gas performs work on the piston as 2300 J of heat are added from the surroundings. During the process, the temperature of the gas decreases by 45K. How much work does the gas perform?

The choices:

A. 1400J, B. 3200J, C. 4400J, D. 5000J, E. 6000J

The diagram given was pretty simple, showing a gas in a chamber pushing against a piston and an arrow going into the gas to show the energy added.

This may seem easy to some, but I am baffled by this question most likely because of my not-so-extensive knowledge in the topic of gases, work done, and high-school thermodynamics.

A thing I tried: I tried to find the average energy of each atom in the ideal gas when 2300J was added to it, and then finding the temperature from that (I assumed that 2300J was added to the ideal gas when was at 0K). There was already one problem: The temperature increase when 2300J is added is only 38.5K (I may be wrong), and that is less than 45K, the temperature that it dropped to. Nonetheless, I found that the energy increased by 391.6624J (which didn't match any of the choices, and wasn't even a decrease in energy).

If anyone could help me point out the correct answer with a brief explanation of the steps, that would be extremely helpful. Also, I think the method that I just tried was completely wrong in all ways, but I'm just curious about it as well.

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  • $\begingroup$ Heat is added and temperature decreases? $\endgroup$
    – Adrian Howard
    Commented Feb 12, 2020 at 5:34
  • $\begingroup$ @Adrian Howard. Do you have a problem with that? $\endgroup$
    – Chet Miller
    Commented Feb 12, 2020 at 12:44
  • $\begingroup$ @ChetMiller Sorry, I thought it was a typo, but I see they must mean temp increases THEN decreases as work is done? Thanks $\endgroup$
    – Adrian Howard
    Commented Feb 12, 2020 at 19:00
  • $\begingroup$ @AdrianHoward. No. That's not what the problem statement says; it says that the work and heat transfer take place simultaneously. $\endgroup$
    – Chet Miller
    Commented Feb 12, 2020 at 20:39

1 Answer 1

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Here are some hints to push you on the right track.

First, you don't need to assume anything other than what the problem says (don't assume a temperature of 0K, etc).

Second, you need to use the first law of thermodynamics $$ \Delta U = Q - W $$ where $\Delta U$ is the change of internal energy, $Q$ is the heat added to the gas and $W$ is the work done by the gas.

You know $Q$ and want to calculate $W$. For that, you also need to know $\Delta U$. But for $\nu$ moles of ideal gas, you have $\Delta U = \nu C_V \Delta T$ where $C_V$ is the molar heat capacity at constant volume. There is a famous formula for this heat capacity...

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