# Homework Help: Thermodynamics help

1. Sep 29, 2013

### S_Flaherty

1. The problem statement, all variables and given/known data
The question is: Consider a hypothetical ideal gas with internal energy U = NkTo(T/T0)α+1, where To and α are positive constants. Show that in an adiabatic process, V*exp[(1+1/α)(T/To)α] = constant.

2. Relevant equations
PVγ = constant
γ = Cp/Cv
Cp = Cv + Nk

3. The attempt at a solution
I'm pretty sure that I'm supposed to show that [(1+1/α)(T/To)α] is equal to γ and since PVγ = constant, V*exp[(1+1/α)(T/To)α] = constant. When I try to solve it though I can't get the solution to come out. I differentiate U to get Cv = Nk(1+α)(T/To)1+α. When I plug that into γ I get γ = 1 + 1/[(1+α)(T/To)α]. Either I'm just not simplifying it enough and the answer is correct, or I solved for λ incorrectly, or my equations are incorrect. I don't know which it is though.

2. Sep 29, 2013

### nasu

The equation
PVγ = constant
is not valid for this problem. It follows from the usual internal energy for ideal gas,
U(T)=nCvT

Here you have a different function U(T) and you have to find the relationship between volume and temperature. (for the "usual" ideal gas this will be TVγ-1 = constant )
You can use the first law in conjunction with the equation of state to do this.

3. Sep 29, 2013

### ehild

Is PVγ = constant when Cv depends on T?

ehild

4. Sep 30, 2013

### S_Flaherty

I'm not really sure what you mean, can you explain it more?

5. Sep 30, 2013

### S_Flaherty

I'm guessing it's not, but I don't know what it should be then.

6. Sep 30, 2013

### Staff: Mentor

For an ideal gas,
dU=CvdT
PV=RT
From the first law, for an adiabatic reversible process, how is dU related to PdV?

7. Sep 30, 2013

### S_Flaherty

dU = -PdV, so Cv = -PdV/dT right?

8. Sep 30, 2013

### ehild

U is given as function of T. PV=kNT is valid for the ideal gas, and also the First Law is valid. For an adiabatic process dU=-PdV. Use P=kNT/V, and integrate.

ehild

9. Sep 30, 2013

### S_Flaherty

Ok, so I get U = -kNT(ln(V))

10. Sep 30, 2013

### haruspex

No, that would not follow, so I don't think you want to show that [(1+1/α)(T/To)α] is equal to γ. Instead, try raising V*exp[(1+1/α)(T/To)α] to the power of γ, using the expression for γ that you derived.

11. Sep 30, 2013

### S_Flaherty

So that makes (V*exp[(1+1/α)(T/To)α])1+1/(1+α)(T/To)α

I'm not sure what to do with that

12. Sep 30, 2013

### nasu

Well, I don't understand what part you don't understand.

But the idea is "forget gamma". And "forget pv^gamma". Does not apply here.

1. From first law applied to adiabatic process you have:
dU=pdV
You have U(T) so find dU.

2. You have PV=nRT so you can eliminate p on the right hand side:
pdV= nRTdV/V

So you will have an equation relating V and T. Integrate (after separating variables) and you'll find that exponential relationship.

13. Sep 30, 2013

### Staff: Mentor

You have the equation of U as a function of T, and you know know that
$$C_v=\frac{\partial U}{\partial T}$$
Just differentiate the equation for U with respect to T, and write
$$dU=C_vdT=\frac{\partial U}{\partial T}dT=-PdV$$
Then, just substitute the ideal gas law for P, and integrate.

14. Sep 30, 2013

### nasu

This is a "hypothetical" ideal gas.
Cv=∂U/∂T is valid for a "real" ideal gas.

The whole point here is that U(T) is not given by
dU=CvdT but by that other, more complicated formula.
If he does what you suggest he'l get just the usual
$$TV^{\gamma -1 }= constant$$ and not the formula required by the problem.

But the method will work. This is what I tried to explain as well.
Just use
$$dU=Nk(\alpha +1) (T/T_0)^{\alpha} dT$$.

There is no need to introduce Cv or gamma.

15. Sep 30, 2013

### Staff: Mentor

I have to disagree. Cv=∂U/∂T follows simply from the definition of heat capacity,
$$C = \frac{Q}{\Delta T}$$
by considering a constant volume (hence $W=0$), without invoking an ideal gas.

16. Sep 30, 2013

### nasu

Did I say anything that seem to contradict your statement? I just meant just that you don't need Cv to solve the problem. It does not appear in this problem.
Oh, I see. I used partial derivatives.

I meant that dU=CvdT may not apply to other systems.
It is valid only for some systems, like ideal gas in the "proper" definition.

So dU=Nk(α+1)(T/T0)^α dT
You don't need to define or use a specific heat to solve the problem.
Sorry for the confusion.

Last edited: Sep 30, 2013
17. Sep 30, 2013

### Staff: Mentor

This is exactly what I was suggesting. I brought the heat capacity into the picture because I felt the OP would feel more comfortable with it. For this particular ideal gas, Cv is not independent of temperature, but is given by:
$$C_v=Nk(\alpha +1) (T/T_0)^{\alpha}$$
Are you uncomfortable with an ideal gas heat capacity that varies with temperature. A temperature-dependent heat capacity is part of the definition of an ideal gas that we engineers use.

18. Sep 30, 2013

### nasu

Is this a question?
I don't feel any discomfort about temperature variation of Cv or about Cv in general. Even Cp it's bearable, despite all these pressure variations.

19. Sep 30, 2013

### Staff: Mentor

Oops. I left out the question mark. Thank you for serving as the grammar police enforcer.

Getting back to the thread, I think we are (and were) totally in agreement on how this problem should be solved. Of course, for an ideal gas, Cp is also a function only of temperature.
Chet

20. Sep 30, 2013

### nasu

I agree that we are in agreement.
It was not intended as police work. Just curious.

21. Oct 1, 2013

### Staff: Mentor

Yes, you did, which is why I wanted to point it out. You said:
I don't see how to read this other than Cv=∂U/∂T is valid only for a "real" ideal gas, not for this "hypothetical" ideal gas. This statement is not correct, as Cv=∂U/∂T is universally valid, except if a phase transition occurs. This may not have been what you were thinking when you wrote that, but I wanted to make things clear.

22. Oct 1, 2013

### ehild

U is given in the OP, and it is explicitly independent on the volume, it is function of T only. But V depends on T. You have to find the relationship between T and V in an adiabatic process, when dU=-PdV. From here, you get a differential equation relating V and T, that you have to integrate. No need to mix gamma in.

ehild

Last edited: Oct 1, 2013
23. Oct 1, 2013

### nasu

Yes, I realized that.
As I already said in my post.