# Thermodynamics help

by S_Flaherty
 P: 75 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.
P: 1,969
 Quote by 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.
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.
HW Helper
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P: 10,523
 Quote by 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.

Is PVγ = constant when Cv depends on T?

ehild

P: 75
Thermodynamics help

 Quote by 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.
I'm not really sure what you mean, can you explain it more?
P: 75
 Quote by ehild Is PVγ = constant when Cv depends on T? ehild
I'm guessing it's not, but I don't know what it should be then.
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PF Gold
P: 5,177
 Quote by S_Flaherty I'm not really sure what you mean, can you explain it more?
For an ideal gas,
dU=CvdT
PV=RT
From the first law, for an adiabatic reversible process, how is dU related to PdV?
P: 75
 Quote by Chestermiller For an ideal gas, dU=CvdT PV=RT From the first law, for an adiabatic reversible process, how is dU related to PdV?
dU = -PdV, so Cv = -PdV/dT right?
 HW Helper Thanks P: 10,523 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
P: 75
 Quote by 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
Ok, so I get U = -kNT(ln(V))
Homework
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P: 9,791
 Quote by S_Flaherty κ 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.
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.
P: 75
 Quote by 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.
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
P: 1,969
 Quote by S_Flaherty I'm not really sure what you mean, can you explain it more?
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.
 Sci Advisor HW Helper Thanks PF Gold P: 5,177 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.
P: 1,969
 Quote by Chestermiller You have the equation of U as a function of T, and you know know that $$C_v=\frac{\partial U}{\partial T}$$ .
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.
PF Gold
P: 1,344
 Quote by nasu This is a "hypothetical" ideal gas. Cv=∂U/∂T is valid for a "real" ideal gas.
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.
P: 1,969
 Quote by DrClaude 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.
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.
HW Helper
Thanks
PF Gold
P: 5,177
 Quote by 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.
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.
P: 1,969
 Quote by Chestermiller 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.
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.

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