Why doesn't ΔU = 0 (U3= U1) in this problem, where pV = constant?

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The discussion centers on a thermodynamics problem where pV = constant, leading to the assumption of an isothermal process, yet the internal energy values U1 and U3 are not equal. It is highlighted that the lack of specification regarding the gas being ideal is crucial, as it affects the applicability of the equation ΔU = 0. Participants suggest solving the problem by calculating the total work done in the cycle using the first law of thermodynamics, without relying on ideal gas equations. The work for the segment where pV is constant is derived through integration, resulting in W3-1 = PV*ln(V1/V3). The conversation emphasizes the importance of recognizing the processes involved, including isochoric and isobaric processes, to accurately determine heat transfer and work done.
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EDIT: My bad, this is from my professors lecture, but I assume this question is more suitable for the homework section of this forum

The problem gives that pV = constant, and thus is assumed to be an isothermal process. However, the given U1 and U3 are not equal to each other. Is it because it is not specified the gas is ideal that we ignore that ΔU = 0 ?

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Yes I think that's the catch, it is not specified that the gas is ideal.

You can solve this problem, without assuming that gas is ideal and without using ##PV=nRT## (or ##\Delta U=nC_V\Delta T##). Just calculate the total work done in the cycle and then use 1st law of thermodynamics for the cycle.

For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$

P.S The solution I see to this problem , doesn't use anywhere the given values of ##U_1,U_3##.
 
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Delta2 said:
Yes I think that's the catch, it is not specified that the gas is ideal.

You can solve this problem, without assuming that gas is ideal and without using ##PV=nRT## (or ##\Delta U=nC_V\Delta T##). Just calculate the total work done in the cycle (as the total area in a PV diagram that is enclosed by the cycle) and then use 1st law of thermodynamics for the cycle.

For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$
Okay cool thank you! So it would thus just be this process-> C = constant = PV, and thus you would integrate 1/V dV from V1 to V3 (leaving c=pv outside of integral) to get W3-1 = PV*ln(V1/V3), where PV = P1V1 or P3V3
 
grotiare said:
Okay cool thank you! So it would thus just be this process-> C = constant = PV, and thus you would integrate 1/V dV from V1 to V3 (leaving c=pv outside of integral) to get W3-1 = PV*ln(V1/V3), where PV = P1V1 or P3V3
Yes that's correct. And also very easily you can calculate the work from ##1 \to 2## (isochoric process) and ##2\to 3## (isobaric process).
 
grotiare said:
The problem gives that pV = constant, and thus is assumed to be an isothermal process.
Wrong assumption.

You have 2 processes - one constant volume, one constant pressure - for which you can calculate the pressure-volume work. (part b)

Then you use the first law to calculate the heat transfer. (part c)
 
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