# Thermodynamics, Callen's problem, it doesn't make sense to me

1. Mar 17, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Two moles of a particular single-component system are found to have a dependence of internal energy U on pressure and volume given by $U=APV^2$ (for N=2) where $A=10cm ^{-3}$. Note that doubling the system doubles the volume, energy, and the mole number but leaves the pressure unaltered, and write the complete dependence of U on P, V and N for arbitrary mole number. Answer: $U=BPV^2/N$ where $B=20 cm ^3$.

2. Relevant equations
Not sure there are any. To be found.

3. The attempt at a solution
First I tried to understand the problem and makes sense of the given expression $U=APV^2$. I checked out the units in SI units and instead of joule I reach $Jm^2$. So that this expression doesn't make sense to me, I cannot even proceed further.
The book is considered of one of the most cited apparently for thermodynamics so I'm sure the book is right and I'm missing something. I just don't know what I'm missing, any help is appreciated.

2. Mar 17, 2012

### vela

Staff Emeritus
How'd you get that?

3. Mar 17, 2012

### fluidistic

By forgetting to divide the force by the area in the pressure expression.

4. Mar 17, 2012

### fluidistic

I don't know how to reach the answer. Attempt:
Let $V_0$ be the volume when $n=2$. So that $V(n)=\frac{V_0 n}{2}$ (linear relation). P is constant. I do the same relation for $U(n)$: $U(n)=\frac{U_0 n}{2}$.
The relation they provide becomes $U_0=APV_0^2$. Written with my terms I get $U(n)=\frac{2APV^2(n)}{n}=\frac{APnV_0 ^2}{2}$.
Something is definitely wrong, I shouldn't be able to get a dependence of U only in terms of n.
I've tried other things but reached nothing like the result.

5. Mar 18, 2012

### rude man

Last edited: Mar 18, 2012
6. Mar 18, 2012

### fluidistic

Thank you for your help. But I think that's what I've done in my last post. I don't see anything wrong with my answer.
When n=2 I get the given relation. When I duplicate n, I get twice the same energy and the volume duplicates too; the pressure remaining the same just as it should. Yet my answer differs from the one given (worth apparently $\frac{200PV^2}{An}$ if I write it in terms of A rather than B.)

7. Mar 18, 2012

### rude man

Last edited: Mar 18, 2012
8. Mar 18, 2012

### fluidistic

Ok rude man, thanks for helping me. I am the OP in case you didn't notice :rofl:
If my answer $U(n)=\frac{2APV^2(n)}{n}$ is right and the answer provided by the textbook is (written in terms of A instead of B) $\frac{200PV^2}{An}$ which is right by definition, how would one shows they are equivalent?
To me, this would mean that 2A=200/A. Wait, that means A=10 (I didn't check out the units though) which is... RIGHT!
Wow.
Problem solved!

9. Mar 19, 2012

### rude man

Yep, and my apologies for thinking you were not the OP!