# Thermal Physics problem: Van der Waals

1. Nov 22, 2013

### mmedrano8

1. Find an expression that relates T and V in a process in which the entropy S is constant, by doing the following:

a) From thermodynamic identity dS=(1/t)dU + (p/v)dV
find the expression for dS as a function of T and V (and dT and dV).

b) Integrate the expression found in (a) along the line of constant S.

2. Relevant equations
I tried to use the equation of state for a Van der Waals gas

p=(NkT)/V-a(N^2/V^2)

a is constant
its energy as a function of temperature and volume is given by
U(T,V)=2NkT-a(N^2/V)

If you could give any hints that would be great!

Last edited: Nov 22, 2013
2. Nov 23, 2013

### rude man

That equation is dimensionally inconsistent and therefore wrong.

But it's just one letter aaway from right so is probably a typo. Anyway, fix it.

Are you allowed to assume an ideal gas?

3. Nov 24, 2013

### mmedrano8

I took a a snapshot of the problem to be more specific

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4. Nov 24, 2013

### TSny

As noted by rude man, the equation given for dS is incorrect. Since the equation is just a simple rearrangement of the first law (with dQ = T dS), you can see how to fix it.

To proceed with the problem, you will need to use the expression for U and find it's differential dU in terms of dT and dV.

5. Nov 24, 2013

### rude man

OK.
You're given U(T,V) and dS = 1/T dU + p/V dV
so taking dU is just basic calculus. N, k and a are constants.

Warning: I've carried this thru to dS = 0 in terms of T and V but I wind up with an expression that I can't integrate (tried separation of parts, couldn't do it). Someone else help!

Last edited: Nov 24, 2013
6. Nov 24, 2013

### mmedrano8

ok so thermodynamic identity --> dS=(1/T)dU + (p/T)dV

7. Nov 24, 2013

### TSny

Good.

What does this expression for dS become after substituting for dU and simplifying?

8. Nov 24, 2013

### rude man

Hey, how come everyone except I caught the misstatement in the problem? Now I gotta start over ... :surprised:

I hope at least U(T,V) is given correctly. Someone else want to check dimensions?

Still can't seem to separate variables though ...

9. Nov 24, 2013

### mmedrano8

ok dU=(∂U/∂T)dT + (∂U/∂V)dV --> dU = 2NKdT + aN^2(1/V^2)dV

I plug in P and dU into dS = (1/T)dU + (P/T)dV

I get...

dS = (1/T)[2NkdT + a(N^2)/(V^2)dV] + (NKT/V - (aN^2)/(V^2))/T)dV

please correct me if im wrong but since S is constant, then dS equals 0,
therefore the lefthand side is 0, we seperate the varibales for dT and dV and simplify
i get
-2dT=(T/V)dV
integrate both parts and solve for Vf --> Vf=Vi/16

Thanks so much guys!!!

10. Nov 24, 2013

### rude man

Congartulations! Would you (or TSny or whoever) mind posting how you did the separation of variables? I couldn't do it.

EDIT: never mind, I hadn't noticed two terms canceling each other, leaving only 2 terms to be separated.

Good show!!

Last edited: Nov 24, 2013
11. Nov 24, 2013

### mmedrano8

i forgot a step, -2dT = (T/V)dV --> (-2/T)dT=(1/V)dV
but i seperated varibles by putting all T's and V's on opposite sides of the equation

-(1/T)[2NkdT] = (1/T)[a(N^2)/(V^2) + NKT/V - (aN^2)/(V^2))/T)dV

simplify a bit...
2NkdT = (NKT/V)dV

2dT = (T/V)dV

(2/T)dT = (1/v)dV then integrate both sides ---> 2ln(Tf/Ti) = ln(Vf/Vi) , where Tf = 4Ti , 2ln(4) = ln(Vf/Vi)
then solve for Vf
=)