# The internal energy in isobaric process

1. Homework Statement

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

2. Homework Equations
ΔU= nCvΔT

3. The Attempt at a Solution

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

SteamKing
Staff Emeritus
Homework Helper
1. Homework Statement

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

2. Homework Equations
ΔU= nCvΔT

3. The Attempt at a Solution

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

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

Chestermiller
Mentor
1. Homework Statement

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

2. Homework Equations
ΔU= nCvΔT

3. The Attempt at a Solution

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

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

Chestermiller
Mentor
actually its not mentioned in the question about procees but the pressure is constant
As I said, what is the exact problem statement?

As I said, what is the exact problem statement?
When volume changes from V to 2V at constant pressure then the change in internal enrgy will be?

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

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

TSny
Homework Helper
Gold Member
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##?

Chestermiller
Mentor
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##?
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

TSny
Homework Helper
Gold Member
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##
If the pressure remains constant, I don't see how the process could be either adiabatic or isothermal.

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

TSny
Homework Helper
Gold Member
Isothermal is easy. You just add enough heat to match the work that was done.
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.

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.

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.

Chestermiller
Mentor
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.
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/

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.
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

TSny
Homework Helper
Gold Member
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/
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.

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

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

Chet

ehild
Homework Helper
1. Homework Statement

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

2. Homework Equations
ΔU= nCvΔT

3. The Attempt at a Solution

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

ehild
Homework Helper
@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.

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TSny
Chestermiller
Mentor
@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.
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

TSny
ehild
Homework Helper
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...

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finally i got the solution

ΔU= nCvΔT
using PΔV = nR ΔT
ΔT= PΔV/nR = PV/nR
So ΔU= nCv (PV/nR)
and when we replace R with Cp-Cv we get the answer

ehild
Homework Helper
finally i got the solution

ΔU= nCvΔT
using PΔV = nR ΔT
ΔT= PΔV/nR = PV/nR
So ΔU= nCv (PV/nR)
and when we replace R with Cp-Cv we get the answer
Well done!