# Coefficients of expansion and compressibility

1. Jan 14, 2010

### Phyisab****

1. The problem statement, all variables and given/known data
I need to derive the following relation:

$$\alpha_{S}=\alpha_{P}\frac{1}{1-\frac{\kappa_{T}}{\kappa_{S}}}$$

2. Relevant equations

Hopefully you can see that my notation |P means at constant pressure, I could not find a better way to do this, any ideas?

$$\alpha_{S}=\frac{1}{V}\frac{\partial V}{\partial T} |P$$

$$\alpha_{P}=\frac{1}{V}\frac{\partial V}{\partial T} |S$$

$$\kappa_{T}=\frac{1}{V}\frac{\partial V}{\partial P} |T$$

$$\kappa_{S}=\frac{1}{V}\frac{\partial V}{\partial P} |S$$

3. The attempt at a solution

I have no intuition about this problem, so i have just been trying everything I can think of and nothing works. I think the first step is pretty obvious:

$$\alpha_{s}=\frac{\alpha_{P}}{\frac{\partial V}{\partial T}|P}\frac{\partial V}{\partial T}|S$$

after this it just seems like a guessing game, trying to apply the proper identity to lead me to the answer. Note that I used one of the Maxwell relations somewhere in here. Here is my last ditch effort. I'm pretty sure the identity I made up is not true, but this seems to have gotten me closest to the answer, and it would take me days to type up all my false leads.

$$dV = \frac{\partial V}{\partial T}|P dT + \frac{\partial V}{\partial P}|T dP$$

From here, I made the almost certainly false conclusion that

$$\frac{\partial V}{\partial T}|S = \frac{\partial V}{\partial T}|P + \frac{\partial V}{\partial P}|T\frac{\partial P}{\partial T}|S$$

Plugging this back into my first step gives:

$$\alpha_{s}= \frac{\alpha_{P}}{1 + \frac{V\kappa_{T}\frac{\partial P}{\partial T}|S}{\frac{\partial V}{\partial T}|S}}$$

Which seems very close to me, but I still can't finish it off and I'm pretty sure I cheated to get there anyway. Any help would be so greatly appreciated, this has got to be a pretty simple problem and it is driving me insane!

2. Jan 14, 2010

### Mapes

Your "almost certainly false conclusion" is fine; you just differentiated both sides with respect to T at constant S. (Try working it through carefully with the chain rule to prove this to yourself, noting that lone differential terms like dT can be assumed to be negligible when compared to partial derivatives.) Try proceeding from there, using partial derivatives and Maxwell relations as necessary.

3. Jan 14, 2010

### Phyisab****

That means the following variations on the fundamental identity all apply:

$$\frac{\partial V}{\partial T}|S=\frac{\partial V}{\partial T}|P+\frac{\partial V}{\partial P}|T\frac{\partial P}{\partial T}|S$$

$$\frac{\partial V}{\partial P}|S=\frac{\partial V}{\partial P}|T+\frac{\partial V}{\partial T}|P\frac{\partial T}{\partial P}|S$$

$$0=\frac{\partial S}{\partial T}|P+\frac{\partial S}{\partial P}|T\frac{\partial P}{\partial T}|S$$

And also the Maxwell relations. Are there others I haven't thought of? I just can't see this problem as anything other than an exercise in pure trial and error.

4. Jan 14, 2010

### Mapes

Trial and error at the beginning, but increased savviness and intuition about which identities to use by the end, after working through many problems.

5. Jan 14, 2010

### Phyisab****

Here is where I am now:

$$\alpha_{S}=\frac{\alpha_{P}}{1-\frac{\frac{\kappa_{T}}{\kappa_{S}}\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}}$$

Now all that is left is to show that

$$\frac{\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}$$

is equal to one. No problem right? How deep can this rabbit hole possibly go.

$$\frac{\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}$$

$$=\frac{\left(\frac{\partial V}{\partial P}|T+\frac{\partial V}{\partial T}|P\frac{T}{P}|S\right)\left( \frac{\frac{\partial V}{\partial T}|S-\frac{\partial V}{\partial T}|P}{\frac{\partial V}{\partial P}|T}\right)}{\frac{\partial V}{\partial T}|S}$$

(To Be Continued...)

6. Jan 14, 2010

### Mapes

Surely it is no problem to show that

$$\left(\frac{\partial P}{\partial T}\right)_S \left(\frac{\partial V}{\partial P}\right)_S \left(\frac{\partial T}{\partial V}\right)_S=1$$

!!!

7. Jan 14, 2010

### Phyisab****

Thanks a ton Mapes. But clearly I am not thinking about all this as well as I could. I knew

$$\left(\frac{\partial P}{\partial T}\right)|V \left(\frac{\partial V}{\partial P}\right) |T \left(\frac{\partial T}{\partial V}\right)|P=1$$,

which is clearly related to what you just said. From what should I try to derive this relation?

8. Jan 14, 2010

### Phyisab****

Nevermind I got it thanks again mapes.