# Solving for λ in an Adiabatic Process: Applying the Ideal Gas Equation"

• S_Flaherty
In summary: I don't understand what you are trying to do. So that makes (V*exp[(1+1/α)(T/To)α])1+1/(1+α)(T/To)α
S_Flaherty
Thermodynamics - adiabatic process

## Homework Statement

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.

PVγ = constant
γ = Cp/Cv
Cp = Cv + Nk

## 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.

S_Flaherty said:

## Homework Statement

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.

PVγ = constant
γ = Cp/Cv
Cp = Cv + Nk

## 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.

S_Flaherty said:

## Homework Statement

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.

PVγ = constant
γ = Cp/Cv
Cp = Cv + Nk

## 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

nasu said:
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?

ehild said:
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.

S_Flaherty said:
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?

Chestermiller said:
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?

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

ehild said:
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))

S_Flaherty said:
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.

haruspex said:
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

S_Flaherty said:
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.

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.

Chestermiller said:
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.

nasu said:
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.

DrClaude said:
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.

Last edited:
nasu said:
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.

Chestermiller said:
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.

nasu said:
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.
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

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

nasu said:
Did I say anything that seem to contradict your statement?
Yes, you did, which is why I wanted to point it out. You said:
nasu said:
This is a "hypothetical" ideal gas.
Cv=∂U/∂T is valid for a "real" ideal gas.
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.

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

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:
DrClaude said:
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.

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

## 1. How do you solve for λ in an adiabatic process?

To solve for λ in an adiabatic process, you would use the ideal gas equation, PV^γ = constant, where γ is the specific heat ratio. You would also need to know the initial and final values of pressure and volume.

## 2. What is the ideal gas equation?

The ideal gas equation, also known as the ideal gas law, is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. It describes the relationship between the four variables for an ideal gas.

## 3. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no transfer of heat between the system and its surroundings. This means that the change in internal energy is equal to the work done on the system.

## 4. How do you apply the ideal gas equation to an adiabatic process?

To apply the ideal gas equation to an adiabatic process, you would use the formula PV^γ = constant, where γ is the specific heat ratio. This equation is derived from the first law of thermodynamics and is used to determine the relationship between pressure and volume in an adiabatic process.

## 5. What is the specific heat ratio?

The specific heat ratio, also known as the adiabatic index or ratio of specific heats, is a thermodynamic property that relates the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv). It is denoted by the symbol γ and is used in the ideal gas equation for adiabatic processes.

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