# Thermal physics

1. Apr 25, 2008

### billybob5588

[SOLVED] thermal physics

1. The problem statement, all variables and given/known data
the differential of enthalpy is given by dH=TdS-VdP

2. Relevant equations
Beginning with the above relationship, find the equation for the differential of the entropy and then show that (d(1/T)/dp)|H=(d(V/T)/dH)|p

3. The attempt at a solution

this is what i got so far...
i rearranged dH=TdS-VdP to make S the subject which gives dS=dH/T+Vdp/T
then since S=S(H,p) and using the definition of the differential of S;
dS=(dS/dH)|p*dH+(dS/dp)|H*dp
i got then 1/T=(dS/dH)|p & V/T=(dS/dp)|H
i dont know what to do next and to how to exactly reach the final term of (d(1/T)/dp)|H=(d(V/T)/dH)|p

2. Apr 25, 2008

### Hootenanny

Staff Emeritus
Welcome to PF billybob,

You're on the right lines and almost there. What do you know about the mixed second partial derivatives of an exact differential?

3. Apr 25, 2008

### billybob5588

exactness condition
i know that they are equal, i thought i would end up somewhere along those lines so..

dS=(dS/dH)|p*dH+(dS/dp)|H*dp

=M(H,p)dH+N(H,p)dp

dM/dp=dN/dH

i see where its getting next bit algebra is getting tricky..

do i cheat and just sub (1/T) for M and and V/T for N?

i dont understand how to show this mathematically nicely

4. Apr 25, 2008

### Hootenanny

Staff Emeritus
Correct.
I'm not quite sure what you doing here, but you know that,

$$\left(\frac{\partial S}{\partial H}\right)_p =\frac{1}{T}$$

$$\left(\frac{\partial S}{\partial P}\right)_H =\frac{V}{T}$$

And you know that,

$$\left[\frac{\partial}{\partial P}\left(\frac{\partial S}{\partial H}\right)_p\right]_H = \left[\frac{\partial}{\partial H}\left(\frac{\partial S}{\partial P}\right)_H\right]_P$$

Can you see what you need to do next (it's a really simple step)?

Last edited: Apr 25, 2008
5. Apr 25, 2008

### billybob5588

subbing in becames..
(d/dp)(1/T)=(d/dH)(V/T)
i cant see how the |H and |p get switched around

6. Apr 25, 2008

### Hootenanny

Staff Emeritus
Correct.
Sorry my bad, I left out the subscripts of what I'm holding constant in my previous post. Does it make more sense now?

7. Apr 25, 2008

### billybob5588

Yes! thanks you so much appreciate it heaps

going to attempt part (b) of the question ill post here if i having difficulties

thanks again!

8. Apr 25, 2008

### Hootenanny

Staff Emeritus

9. Apr 27, 2008

### billybob5588

part (b)

now for part (b) of this question

Using the equation of state for an ideal gas and the definition of enthalpy (H=U+pV), express 1/T and V/T as function of p and H and show that the relationship in part (a) is valid for an ideal gas.

trying to understand what this question means..

i know pV=nRT, do i sub that in enthalpy equation which becomes H=U+nRT?

want 1/T(p,H) and V/T(p,H)

i really don't know where to start for this question and how to approach it.

some hints will be much appreciated;

Last edited: Apr 27, 2008
10. Apr 27, 2008

### Hootenanny

Staff Emeritus
So what you want is

$$\frac{1}{T}=f(p,H)\hspace{2cm}\frac{V}{T}=F(p,h)$$

With respect to the definition of the enthalpy, bearing in mind that we are dealing with an ideal gas (no potential energy), can you write U as a function of T?

11. Apr 27, 2008

### billybob5588

i thought U=0 like you said above, since its an ideal gas which makes H=pV;
i am sorry i don't understand writing U as a function of T since U is 0;

so V=H/p

pV=nRT => p(H/p)=nRT => T=H/nR

that right?

sorry i am completely puzzled by this

12. Apr 27, 2008

### Hootenanny

Staff Emeritus
Not quite. U is the internal energy, which roughly speaking is the sum of the potential energies and kinetic energies of the gas. Now as I said previously, since the gas is ideal, there is no potential energy; but there is still kinetic energy. Now, how is the kinetic energy of an ideal gas related to it's temperature? (HINT: Equipartition theorem).

13. Apr 27, 2008

### billybob5588

the relationship is;
change is kinetic energy means a change in temperature (proportional to each other).

equipartition theorem = 3/2RT

write U as a function of T is is U=3/2RT or U=C*T (where C is a constant)

so H=U+pV

H=CT+pV
that correct? and then make V/T and 1/T the subject?

14. Apr 27, 2008

### Hootenanny

Staff Emeritus
Are you sure it's R? Apart from that it looks good.

15. Apr 27, 2008

### billybob5588

its 3/2kT sorry;

ok so rearranging to make 1/T and V/T the subject i get
1/T=1/(H-pV)
V/T=H/TP - 1/P => for this one im not sure cause V/T is not the clear subject there is a T term in the equation

and now i should plug it into equation in part (a) then solve?

i thank you for your continuing help

Last edited: Apr 27, 2008
16. Apr 27, 2008

### Hootenanny

Staff Emeritus
Better
I think your missing a factor of C here.
For this one, I think it would be must easier to use the ideal gas eqaution and simply find V/T as a function of P.

17. Apr 27, 2008

### billybob5588

sorry my algebra is a bit weak

H=kT+pV where pV=nRT
H=kT+nRT
i dont see how this can help
and here probably alot more worse i know this is wrong but im trying everything.
H=kT+n^2*R^2*T^2/Vp (i got this from p=nRT/V and V=nRT/p and taking the product of this yeilds n^2*R^2*T^2/Vp)

this is embarassing

maths...

18. Apr 27, 2008

### Hootenanny

Staff Emeritus
Your second expression need not be a function of H, what I meant was,

$$PV = nRT \Rightarrow \frac{V}{T} = \frac{nR}{P}$$

Do you follow?

19. Apr 27, 2008

### billybob5588

yes i understand where u got that from but my partial derivative on RHS is [d(V/T)/dH]|p so i need it in terms of H so i can proceed with the partial and proove they are equal

20. Apr 27, 2008

### Hootenanny

Staff Emeritus
You're quite right, I forgot that the partial derivative was with respect to H . So, using the definition of enthalpy and the ideal gas law,

$$H = \left(C+nR\right)T \Rightarrow T = \frac{H}{C+nR}$$

And,

$$PV = nRT \Rightarrow \frac{V}{T} = \frac{nR}{P}$$

Now, you should be able to combine the two and hence form an expression for V/T in terms of P and H.