# Thermal physics

billybob5588
[SOLVED] thermal physics

## Homework Statement

the differential of enthalpy is given by dH=TdS-VdP

## Homework 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

## 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 don't 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

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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?

billybob5588
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?

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 don't understand how to show this mathematically nicely

Staff Emeritus
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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
Correct.
=M(H,p)dH+N(H,p)dp

dM/dp=dN/dH
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)?

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billybob5588
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,

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

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

subbing in becames..
(d/dp)(1/T)=(d/dH)(V/T)
i can't see how the |H and |p get switched around

Staff Emeritus
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subbing in becames..
(d/dp)(1/T)=(d/dH)(V/T)
Correct.
i can't see how the |H and |p get switched around

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

billybob5588
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?

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!

Staff Emeritus
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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!

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;

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Staff Emeritus
Gold Member
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?

billybob5588
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?

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

Staff Emeritus
Gold Member
i thought U=0 like you said above, since its an ideal gas which makes H=pV;
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).

billybob5588
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).

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?

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

billybob5588
Are you sure it's R? Apart from that it looks good.

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 I am 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

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Staff Emeritus
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its 3/2kT sorry
Better
1/T=1/(H-pV)
I think your missing a factor of C here.
V/T=H/TP - 1/P => for this one I am not sure cause
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.

billybob5588
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.

sorry my algebra is a bit weak

H=kT+pV where pV=nRT
H=kT+nRT
i don't see how this can help
and here probably a lot more worse i know this is wrong but I am 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...

Staff Emeritus
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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?

billybob5588
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?

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

Staff Emeritus
Gold Member
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
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.

billybob5588
ok i got it, thanks for your help!

ill bugger off now haha;

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