# I Thermodynamic Process

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1. Jan 19, 2017

### Ian Baughman

What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p1, p2, V1, and V2 is given by the following.

1) WBC=p2(VC-VB)
2) Isothermal from A → B so p1V1 = p2VB so VB=(p1V1)/p2
3) Similarly VC=(p1V2)/p2
4) WBC= p2((p1V2)/p2-(p1V1)/p2)
5) Factor out p1/p2 and simplify the you get;
WBC=p1(V2-V1)​

2. Jan 19, 2017

The isothermal processes have no change in internal energy. for an ideal gas. Much heat must be added to the system in the isobaric expansion for two reasons: 1)work is being done as it expands 2) It is going to a higher temperature state. In the isobaric compression, much heat must be removed from the system for two reasons 1) The system is going to a lower temperature 2) Work is being done on the system. The energy from this work must also be removed by removing heat. I do think your calculations are correct. Thermodynamic problems can sometimes be tricky, but I believe you analyzed it correctly.

3. Jan 20, 2017

It should also be noted that you can readily compute $\Delta Q=\Delta W$ for each of the two isotherms, since each has $\Delta U=0$. $\\$ Meanwhile $\Delta U$ for each of the isobars will be $\Delta U=\frac{3}{2}nR \Delta T$ for an ideal monatomic gas, and $\Delta U=\frac{5}{2}nR \Delta T$ for an ideal diatomic gas (where $\Delta T=\Delta (PV)/(nR)$). You can thereby readily compute $\Delta Q$ for each of the isobars, if you know the type of gas. $\\$ I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?

Last edited: Jan 20, 2017
4. Jan 20, 2017

### Ian Baughman

Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?

5. Jan 20, 2017

Yes. And if you will notice, for the two isotherms $\Delta U=0$ and for the transitions on the two isobars, $\Delta U_1=-\Delta U_2$. $\\$ Editing... perhaps we haven't proved yet that $\Delta (PV)$ is equal and opposite for these two cases...but that is obvious because on one isotherm, $PV=nRT_1$ and on the other isotherm $PV=nRT_2$.Thereby $\Delta (PV)=nR \Delta T$ and $\Delta U=\frac{3}{2} nR \Delta T$ or $\Delta U=\frac{5}{2} nR \Delta T$.

6. Jan 21, 2017

### Staff: Mentor

Yes. So....

And yes, since this a thermodynamic cycle, the change in internal energy for the working fluid has to be zero.