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

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
  • #1
fluidistic
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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 [itex]U=APV^2[/itex] (for N=2) where [itex]A=10cm ^{-3}[/itex]. 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: [itex]U=BPV^2/N[/itex] where [itex]B=20 cm ^3[/itex].


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 [itex]U=APV^2[/itex]. I checked out the units in SI units and instead of joule I reach [itex]Jm^2[/itex]. 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.
 
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  • #2
fluidistic said:
First I tried to understand the problem and makes sense of the given expression [itex]U=APV^2[/itex]. I checked out the units in SI units and instead of joule I reach [itex]Jm^2[/itex].
How'd you get that?
 
  • #3
vela said:
How'd you get that?
By forgetting to divide the force by the area in the pressure expression. :smile:
 
  • #4
I don't know how to reach the answer. Attempt:
Let [itex]V_0[/itex] be the volume when [itex]n=2[/itex]. So that [itex]V(n)=\frac{V_0 n}{2}[/itex] (linear relation). P is constant. I do the same relation for [itex]U(n)[/itex]: [itex]U(n)=\frac{U_0 n}{2}[/itex].
The relation they provide becomes [itex]U_0=APV_0^2[/itex]. Written with my terms I get [itex]U(n)=\frac{2APV^2(n)}{n}=\frac{APnV_0 ^2}{2}[/itex].
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
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 [itex]U=APV^2[/itex] (for N=2) where [itex]A=10cm ^{-3}[/itex]. 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: [itex]U=BPV^2/N[/itex] where [itex]B=20 cm ^3[/itex].


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.
 
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  • #6
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 [itex]\frac{200PV^2}{An}[/itex] if I write it in terms of A rather than B.)
 
  • #7
fluidistic said:
I don't know how to reach the answer. Attempt:
Let [itex]V_0[/itex] be the volume when [itex]n=2[/itex]. So that [itex]V(n)=\frac{V_0 n}{2}[/itex] (linear relation). P is constant. I do the same relation for [itex]U(n)[/itex]: [itex]U(n)=\frac{U_0 n}{2}[/itex].
The relation they provide becomes [itex]U_0=APV_0^2[/itex]. Written with my terms I get [itex]U(n)=\frac{2APV^2(n)}{n}=\frac{APnV_0 ^2}{2}[/itex].
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.
 
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  • #8
Ok rude man, thanks for helping me. I am the OP in case you didn't notice :smile:
If my answer [itex]U(n)=\frac{2APV^2(n)}{n}[/itex] is right and the answer provided by the textbook is (written in terms of A instead of B) [itex]\frac{200PV^2}{An}[/itex] 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!
 
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  • #9
fluidistic said:
Ok rude man, thanks for helping me. I am the OP in case you didn't notice :smile:
If my answer [itex]U(n)=\frac{2APV^2(n)}{n}[/itex] is right and the answer provided by the textbook is (written in terms of A instead of B) [itex]\frac{200PV^2}{An}[/itex] 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! :blushing:
 

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