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

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Discussion Overview

The discussion revolves around a thermodynamics problem involving a process where pV is constant. Participants explore why the change in internal energy (ΔU) is not zero when the initial and final states of the system (U1 and U3) are given as unequal. The conversation touches on concepts related to ideal and non-ideal gases, the first law of thermodynamics, and the calculation of work done in different processes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the lack of specification regarding the gas being ideal is crucial to understanding why ΔU is not zero.
  • Others propose that the problem can be solved without assuming the gas is ideal and without using the ideal gas law or the formula for ΔU.
  • One participant mentions calculating the total work done in the cycle and applying the first law of thermodynamics to find heat transfer.
  • Another participant emphasizes that there are two distinct processes involved: one at constant volume and one at constant pressure, which allows for the calculation of pressure-volume work.

Areas of Agreement / Disagreement

Participants generally agree that the assumption of the gas being ideal is a key factor in the discussion. However, there are competing views on how to approach the problem and the implications of the processes involved, indicating that the discussion remains unresolved.

Contextual Notes

Participants express uncertainty regarding the assumptions made about the gas and the implications for the internal energy change. There is also a lack of consensus on the correct approach to calculating work and heat transfer in the given processes.

grotiare
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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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