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

Gold Member

## Homework Statement

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 $B=20 cm ^3$.

## Homework Equations

Not sure there are any. To be found.

## 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.

Staff Emeritus
Homework Helper
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 $Jm^2$.
How'd you get that?

Gold Member
How'd you get that?
By forgetting to divide the force by the area in the pressure expression. Gold Member
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.

Homework Helper
Gold Member

## Homework Statement

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 $B=20 cm ^3$.

What's confusing is the statement that doubling the system doubles U, n and V, yet the formula seems to suggest that doubling the system quadruples U, since it seems that if V' = 2V, so U' = 10p(2V)^2 = 40pV2 = 4U. But this is not what the problem states: it states that U just doubles & we have to accommodate that fact:

Oops! I nearly did it again, giving the complete solution.

Last edited:
Gold Member
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.)

Homework Helper
Gold Member
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.[/QUOTE

EDIT! EDIT! EDIT!

fluidistic - I don't know how to tell you this, but your answer was always correct:
U(n) = {2ApV2(n)}/n is correct and agrees with the answer the OP gave us!

*****************************************************************

I'll give OP a lead: U(n) = (10pn/2)V2(2)
but V(2) = 2V(n)/n so
U(n) = (10pn/2){2V(n)/n}2

You wind up with the given answer.

While I'm at it:it makes sense, does it not, that doubling the volume does not change the internal energy? For example, in a free expansion where no work is done and no heat is added or subtracted.

Last edited:
Gold Member
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!

Homework Helper
Gold Member
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!

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