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

• fluidistic
In summary, the conversation discusses a thermodynamics problem involving a single-component system with a dependence of internal energy on pressure and volume. The given expression for the dependence is U=APV^2, where A is equal to 10cm^{-3} and N is equal to 2. It is noted that doubling the system results in doubling the volume, energy, and mole number, while leaving the pressure unchanged. The complete dependence of U on P, V, and N for an arbitrary mole number is given by U=BPV^2/N, where B is equal to 20 cm^{-3}. The conversation also includes a discussion on how to arrive at this answer, with the final solution being U(n)=(10pn/2)V
fluidistic
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.

fluidistic said:
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?

vela said:
How'd you get that?
By forgetting to divide the force by the area in the pressure expression.

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.

fluidistic said:

## 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:
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.)

fluidistic said:
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:
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!

fluidistic said:
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!

## 1. What is thermodynamics?

Thermodynamics is the branch of science that deals with the study of energy and its transformations in relation to matter. It explains how energy is transferred and converted between different forms, such as heat, work, and internal energy.

## 2. What is Callen's problem?

Callen's problem is a theoretical physics problem that deals with the interpretation of the second law of thermodynamics. It relates to the concept of entropy and how it can decrease in certain systems, which goes against the traditional understanding of the second law.

## 3. What are the implications of Callen's problem?

The implications of Callen's problem are still a subject of debate among scientists. One possible implication is that it could lead to a better understanding of the fundamental principles of thermodynamics and potentially challenge our current understanding of the laws of thermodynamics.

## 4. How can we solve Callen's problem?

There is no definitive solution to Callen's problem at this time. Some scientists have proposed new theories and interpretations of the second law of thermodynamics, while others argue that the problem may not actually exist. Further research and experimentation are needed to fully understand and potentially solve Callen's problem.

## 5. Why is it important to understand thermodynamics and Callen's problem?

Understanding thermodynamics and Callen's problem is crucial for many areas of science and technology, such as energy production, materials science, and environmental studies. It also allows us to make accurate predictions and calculations about energy transformations and the behavior of complex systems.

Replies
1
Views
901
Replies
5
Views
946
Replies
7
Views
2K
• Thermodynamics
Replies
23
Views
1K
• Thermodynamics
Replies
19
Views
1K
Replies
2
Views
2K
Replies
1
Views
977
Replies
1
Views
1K
• Chemistry
Replies
1
Views
966
• Thermodynamics
Replies
3
Views
819