# The internal energy in isobaric process

1. Jan 9, 2016

### Jiya

1. The problem statement, all variables and given/known data

When volume changes from V to 2V at constant pressure then the change in internal enrgy will be?

2. Relevant equations
ΔU= nCvΔT

3. The attempt at a solution

I have used the equation ΔU= nCvΔT but the answer is PV/γ-1

2. Jan 9, 2016

### SteamKing

Staff Emeritus
Cv is the specific heat at constant volume. You don't have constant volume, since it changes from V to 2V.

You need the specific heat at constant pressure here.

https://en.wikipedia.org/wiki/Isobaric_process

3. Jan 9, 2016

### Staff: Mentor

For an ideal gas, $\Delta U=C_v\Delta T$ for all process paths (always), even if the volume is not constant.

4. Jan 9, 2016

### Staff: Mentor

I'm just guessing that you left out of the problem statement the constraint that the expansion is adiabatic. Did you? What else did you leave out? What is the exact problem statement (not your interpretation)?

Chet

5. Jan 9, 2016

### Jiya

actually its not mentioned in the question about procees but the pressure is constant

6. Jan 9, 2016

### Staff: Mentor

As I said, what is the exact problem statement?

7. Jan 9, 2016

### Jiya

When volume changes from V to 2V at constant pressure then the change in internal enrgy will be?

8. Jan 9, 2016

### Staff: Mentor

OK there is not enough information specified to solve this problem.

9. Jan 9, 2016

### TSny

I think there is enough information as long as you assume the gas is ideal.
Jiya, do you know the relation between $\gamma$ and $C_V$?

10. Jan 9, 2016

### Staff: Mentor

Really? That would be very interesting. Consider this:

If the expansion is adiabatic, I get $\Delta U=-W=-P\Delta V=-PV$

If the expansion is isothermal, for an ideal gas, I get $\Delta U=0$

Chet

11. Jan 9, 2016

### TSny

If the pressure remains constant, I don't see how the process could be either adiabatic or isothermal.

12. Jan 9, 2016

### Staff: Mentor

Isothermal is easy. You just add enough heat to match the work that was done. Constant pressure involves starting at equilibrium at a higher pressure than P, and then suddenly dropping the applied external force per unit area to P and letting the gas expand to twice the volume while controlling the external force per unit area to the constant value of P.

13. Jan 9, 2016

### TSny

From the ideal gas law: $T = \frac{PV}{nR}$. If P remains constant while V increases, then T must increase. Therefore the process can't be isothermal, as I see it.

I'm interpreting the problem to mean that the pressure, P, of the gas remains constant during the process. Perhaps that's an assumption that I shouldn't be making? I would also interpret "adiabatic" to imply the ususal $PV^\gamma = const$ , thus P could not remain constant while V increases.

I have a feeling that you have a more sophisticated way of looking at this than my freshman physics approach.

14. Jan 9, 2016

### Staff: Mentor

It can be done with an irreversible expansion, where you suddenly drop the externally applied pressure to half its original equilibrium value, and keep the gas in contact with a constant temperature reservoir at the original temperature, so that the final volume is twice the original value. In Thermo, this counts as a constant pressure process, even though there is a discontinuous change in applied pressure initially. Also, in an irreversible expansion, the ideal gas law cannot be applied to the gas, because the pressure and temperature are not even uniform within the gas and, in addition, the pressure at the interface includes a contribution from viscous stresses, which are related, not to the volume, but to the rate of change of volume. For more details on this, please see my Physics Forums Insights article at this link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

TSny!!!! I can't believe you are only a freshman. From many of your postings, I was sure you were a PhD. Very impressive.

Again, for an irreversible adiabatic expansion, the ideal gas reversible expansion equation $PV^\gamma = const$ cannot be applied. But, here too, it is possible to drop the pressure to a lower value initially, and then allow the gas to equilibrate at twice the original volume. However, here the analysis is a little more complicated. One would need to figure out how much lower to drop the pressure in order for the final volume to be twice the initial. And the behavior would of course also involve a decrease in temperature.

Chet

15. Jan 9, 2016

### TSny

OK, I knew there must have been something I was missing. I was assuming the process was reversible (or at least quasi-static) so that pressure of the gas would be well-defined at each step of the process. If so, then you get the stated answer. So, I take it that the reason you said that there was not enough information given is that they did not specify that the process was quasi-static.

LOL (I'm not sure, but I think a felt a tug on one of my legs.) If i were still a freshman I'd be the oldest one in the world. I just meant I was approaching this problem like a freshman. I'm in the stage of life where mental faculties are definitely declining. I often don't trust my answers here on the forum and actually enjoy being challenged. I'm still learning.

16. Jan 9, 2016

### Staff: Mentor

I hadn't thought of the possibility that it could be quasi-static (reversible). Duh!!!!! You're absolutely right. I stand corrected.

Chet

17. Jan 10, 2016

### ehild

And how much is ΔT if the pressure is the same in the initial and final states and the volume becomes twice the initial?
Remember, what γ is. You should see that your result is the same as the given one, PV/(γ-1) ( you left out the parentheses!) .

18. Jan 10, 2016

### ehild

@chet, the problem gives that the final volume is twice the initial one, while the pressure is the same. It does not matter how the final state was obtained. U, T, V, and P are state variables. Assuming an ideal gas (why not on this level) the final temperature and the change of the internal energy from it are easy to obtain. The official solution expressed R by Cp and Cv.

Last edited: Jan 10, 2016
19. Jan 10, 2016

### Staff: Mentor

Thanks Ehild.

The source of my confusion on this was my Thermo experience in which we frequently analyze what we call "constant pressure irreversible expansions." This is where the gas is initially at equilibrium at a certain pressure and, at time zero, we drop the external pressure to a lower value and allow the gas to expand against this lower pressure until it re-equilibrates. Even though the constant pressure that the gas expands against differs from the initial equilibrium pressure, all the work is done against this constant pressure, and we call this a constant pressure irreversible expansion. Here is an example of such a problem that I recently saw on Chemistry Stack Exchange:

A $\mathrm{5.00\ L}$ sample of CO2 at $800 \ \mathrm{kPa}$ underwent a one-step (irreversible) adiabatic expansion against a constant external pressure of $100\ \mathrm{kPa}$. The initial temperature of the gas was $300\ \mathrm{K}$. Give equations (in terms of V and T, the final volume and temperature of the gas) for ΔU, q and w. Calculate the final volume and temperature of the gas.

In a recent Physics Forums Insights article, I discussed the fundamental differences between reversible and irreversible expansions (using a spring-damper analogy), and presented analytical results for a constant pressure irreversible gas expansion to illustrate the analogy. Here is a link to that article: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

Chet

20. Jan 10, 2016

### ehild

I see your point, but I think "When volume changes from V to 2V at constant pressure" meant the pressure of the gas was the same in the initial and final states and the difference of internal energy between these states was the question. My mind works according to high-school level...