# [Thermodynamics] Temperature change during cooling of a gas

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1. Nov 16, 2014

### mef51

1. The problem statement, all variables and given/known data
One method for cooling a gas is adiabatic throttling (Joule-Thomson Experiment). Another method is a reversible adiabatic expansion. Show that if the initial and final pressures are the same, the difference in temperature obtained by the second method is always higher.
Hint: Express $\frac{\partial T}{\partial P}$ in each process as a function of the expansivity $\beta$, the isothermal compressibility $\kappa$ and $c_P$ and compare the results.

2. Relevant equations
The question has to do with thermodynamic potentials.
$$\beta=\frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_{P} \\ \kappa=-\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_{T} \\ c_{P}=\frac{T}{n}\Big(\frac{\partial S}{\partial T}\Big)_{P}$$
Cyclic Relation:
$$\Big(\frac{\partial T}{\partial P}\Big)_{S}\Big(\frac{\partial P}{\partial S}\Big)_{T}\Big(\frac{\partial S}{\partial T}\Big)_{P}=-1$$
Reciprocal Relation
$$\Big(\frac{\partial x}{\partial y}\Big)_{z}=\Big(\frac{\partial y}{\partial x}\Big)_{z}^{-1}$$

The temperature is a function of $S$ and $P$ so
$$T(P,S) \implies dT=\Big(\frac{\partial T}{\partial S}\Big)_{P}dS+\Big(\frac{\partial T}{\partial P}\Big)_{S}dP$$

The Maxwell Relations are also relevant.

3. The attempt at a solution
Using the cyclic and reciprocal relation and the definitions of $\beta$, $\kappa$ and $c_P$ I can get that, for the second process
$$\Big(\frac{\partial T}{\partial P}\Big)_{S}=\frac{\beta VT}{nc_{p}}$$
To get the change in temperature I integrate the differential. Since the second process is reversible $dS=0$ so
$$\Delta T=T_{f}-T_{i}=\int_{P_{i}}^{P_{f}}\Big(\frac{\partial T}{\partial P}\Big)_{S}dP$$
But since the final and initial pressures are the same this change in temperature is always 0! How can it ever be greater than the temperature difference in the first method? What am I missing here?

2. Nov 16, 2014

### Bystander

Let's insert, "for both processes," right in here, and see how it works.

3. Nov 16, 2014

### mef51

Hm okay well for adiabatic throttling I know that the Joule-Thomson coefficient $\mu$ is positive when a gas is cooling so we know that $\mu \equiv \Big( \frac{\partial T}{\partial P} \Big)_H > 0$.

Using the fact that the adiabatic throttling is a constant-enthalpy process you can show that
$$\Big(\frac{\partial T}{\partial P}\Big)_{H}=\frac{V}{nc_{p}}(\beta T-1)=\frac{\beta VT}{nc_{p}}-\frac{V}{nc_{p}}$$
I notice that $\Big(\frac{\partial T}{\partial P}\Big)_{S}$ is related to this expression and we have
$$\Big(\frac{\partial T}{\partial P}\Big)_{H}=\Big(\frac{\partial T}{\partial P}\Big)_{S}-\frac{V}{nc_{p}}$$

But even with this if I go to do the integral to find $\Delta T$ I still get zero because the final and initial pressure's are the same... $dP=0$!

Since $\frac{V}{nc_p}$ is positive I can see that
$$\Big(\frac{\partial T}{\partial P}\Big)_{S}>\Big(\frac{\partial T}{\partial P}\Big)_{S}-\frac{V}{nc_{p}}=\Big(\frac{\partial T}{\partial P}\Big)_{H}=\mu>0$$

But I don't understand how that would imply the temperature difference being greater. They both seem to be 0

Last edited: Nov 16, 2014
4. Nov 16, 2014

### Staff: Mentor

The intial pressure is the same for both processes, the final temperature is the same for both processes, but the initial temperature is not the same as the final temperature.

5. Nov 16, 2014

### mef51

The initial and the final pressures are the same.

How do you know that? I could equivalently say that the initial temperatures are the same for both processes and the final temperatures are different. Further, it's not even necessary that any of the temperatures are the same. We just want to show that cooling by the second method gives a larger temperature difference than the first method. (EDIT: misread). I know that the initial and final temperatures must be different which is why I'm confused that the integral that gives $\Delta T$ gives 0

Last edited: Nov 16, 2014
6. Nov 16, 2014

### Staff: Mentor

Because you can neither have adiabatic expansion nor throttling at constant pressure. You are misunderstanding the problem, as Bystander also tried to tell you.

7. Nov 16, 2014

### mef51

So the pressure starts and ends at the same point but changes in between?

Last edited: Nov 16, 2014
8. Nov 16, 2014

### Staff: Mentor

How can you do a reversible adiabatic expansion of a gas if the initial and final pressures are the same? How can you throttle a gas adiabatically if the initial and final pressures are the same?

Chet

9. Nov 16, 2014

### mef51

I dunno! On a PV diagram an adiabat with constant pressure doesn't really make sense either.
Let's say that we don't have this condition that the initial and final pressures are the same. Then we'd have that the temperature difference for each method is
$$\newcommand{\pderv}[3]{\Big(\frac{\partial#1}{\partial#2}\Big)_{#3}} \Delta T_{1}=\int_{P_{i}}^{P_{f}}\pderv TPHdP\qquad\Delta T_{2}=\int_{P_{i}}^{P_{f}}\pderv TPSdP$$
which means, since $\pderv TPS>\pderv TPH$
$$\Delta T_{2}>\Delta T_{1}$$

I checked that I didn't copy the question down wrong. Maybe it's a typo

10. Nov 16, 2014

### Bystander

Quod erat demonstrandum.

11. Nov 16, 2014

### mef51

Indeed

12. Nov 17, 2014

### Staff: Mentor

It's not a typo. I've been guilty myself of writing questions with that kind of formulation (in exams, no less). What happens is that the meaning of "the initial and final pressures are the same" is evident in the mind of the person writing the question, to the point of not realizing it can be misconstrued.

13. Nov 17, 2014

### mef51

Ohh haha the lightbulb just clicked.
Perhaps the unambiguous way of phrasing it would be
"Show that if both processes start at an initial pressure $P_i$ and end at a final pressure $P_f$, then the difference in temperature obtained by the second method is always higher."

14. Nov 17, 2014

### Staff: Mentor

Yes. That would have been much, much, much, much better. The original version confused lots of us.

Chet