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Thermodynamics help

by S_Flaherty
Tags: adiabatic, energy, process, thermodynamics
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S_Flaherty
#1
Sep29-13, 09:07 PM
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.
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nasu
#2
Sep29-13, 10:06 PM
P: 1,970
Quote Quote by S_Flaherty View Post
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.
ehild
#3
Sep29-13, 10:32 PM
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Quote Quote by S_Flaherty View Post
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

S_Flaherty
#4
Sep30-13, 10:58 AM
P: 75
Thermodynamics help

Quote Quote by nasu View Post
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?
S_Flaherty
#5
Sep30-13, 11:00 AM
P: 75
Quote Quote by ehild View Post
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.
Chestermiller
#6
Sep30-13, 11:08 AM
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Quote Quote by S_Flaherty View Post
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?
S_Flaherty
#7
Sep30-13, 11:18 AM
P: 75
Quote Quote by Chestermiller View Post
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?
ehild
#8
Sep30-13, 11:23 AM
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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
S_Flaherty
#9
Sep30-13, 11:31 AM
P: 75
Quote Quote by ehild View Post
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))
haruspex
#10
Sep30-13, 11:45 AM
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Quote Quote by S_Flaherty View Post
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.
S_Flaherty
#11
Sep30-13, 12:36 PM
P: 75
Quote Quote by haruspex View Post
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
nasu
#12
Sep30-13, 01:31 PM
P: 1,970
Quote Quote by S_Flaherty View Post
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.
Chestermiller
#13
Sep30-13, 01:57 PM
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You have the equation of U as a function of T, and you know know that
[tex]C_v=\frac{\partial U}{\partial T}[/tex]
Just differentiate the equation for U with respect to T, and write
[tex]dU=C_vdT=\frac{\partial U}{\partial T}dT=-PdV[/tex]
Then, just substitute the ideal gas law for P, and integrate.
nasu
#14
Sep30-13, 02:14 PM
P: 1,970
Quote Quote by Chestermiller View Post
You have the equation of U as a function of T, and you know know that
[tex]C_v=\frac{\partial U}{\partial T}[/tex]
.
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
[tex]TV^{\gamma -1 }= constant[/tex] and not the formula required by the problem.

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

There is no need to introduce Cv or gamma.
DrClaude
#15
Sep30-13, 02:54 PM
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Quote Quote by nasu View Post
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.
nasu
#16
Sep30-13, 03:26 PM
P: 1,970
Quote Quote by DrClaude View Post
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.
Chestermiller
#17
Sep30-13, 07:02 PM
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Quote Quote by nasu View Post
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
[tex]TV^{\gamma -1 }= constant[/tex] and not the formula required by the problem.

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

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:
[tex]C_v=Nk(\alpha +1) (T/T_0)^{\alpha}[/tex]
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.
nasu
#18
Sep30-13, 08:38 PM
P: 1,970
Quote Quote by Chestermiller View Post
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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