Helmholtz Free Energy: Finding C_H-C_M Relation

In summary, the free energy of an incompressible liquid placed in a magnetic field can be written as dA=-SdT-VMdH, and the two heat capacities can be defined as C_H = T (dS/dT)_H and C_M = T (dS/dT)_M. By expressing dS as a function of temperature and magnetic field, the relation for C_H - C_M can be found using the chain rule and Maxwell relations. The final expression is C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H = -\frac{
  • #1
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Homework Statement



The Helmholtz free energy is written as:

[tex] dA=-SdT-pdV-VMdH [/tex]

when an incompressible liquid is placed in a magnetic field H. Thus, the free energy can be written as since [tex]-pdV=0[/tex]:

[tex] dA=-SdT-VMdH [/tex]

The two heat capacities can be defined as:

[tex]C_H = T (dS/dT)_H [/tex] and [tex]C_M = T (dS/dT)_M [/tex]

Where one is at constant H field and the other is at constant magnetization.

I need to find the relation for [tex] C_H - C_M [/tex]

I'm really not sure where to begin for this problem, any advice would be really helpful. Thank you.
 
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  • #2
There's a trick to these problems that's not immediately obvious: express dS as

[tex]dS=\left(\frac{\partial S}{\partial T}\right)_H\,dT+\left(\frac{\partial S}{\partial H}\right)_T\,dH[/tex]

and take the derivative with respect to T at constant M. Know what I mean?
 
  • #3
I'm not sure exactly if I have the entire concept down but here's what I'm thinking:

[tex]C_M=T \frac{1}{dT} \left( \left(\frac{\partial S}{\partial T}\right)_H dT+\left(\frac{\partial S}{\partial H}\right)_T dH \right)_M [/tex]

[tex]C_M=T \left( \left(\frac{\partial S}{\partial T}\right)_H +\left(\frac{\partial S}{\partial H}\right)_T \frac{dH}{dT} \right)_M [/tex][tex]C_H=T \frac{1}{dT} \left( \left(\frac{\partial S}{\partial T}\right)_M dT+\left(\frac{\partial S}{\partial M}\right)_T dM \right)_H [/tex]

[tex]C_H=T \left( \left(\frac{\partial S}{\partial T}\right)_M +\left(\frac{\partial S}{\partial M}\right)_T \frac{dM}{dT} \right)_H [/tex]

thus:

[tex]C_H-C_M[/tex]
[tex]C_H-C_M=T \left( \left( \frac{\partial S}{\partial T} \right)_{M,H} - \left( \frac{\partial S}{\partial T} \right)_{M,H} + \left( \frac{\partial S}{\partial M} \right)_{T.H} \left( \frac{\partial M}{\partial T} \right)_H - \left( \frac{\partial S}{\partial H} \right)_{T.H} \left( \frac{\partial H}{\partial T} \right)_M \right)[/tex]
[tex]C_H-C_M=T \left( \left( \frac{\partial S}{\partial M} \right)_{T.H} \left( \frac{\partial M}{\partial T} \right)_H - \left( \frac{\partial S}{\partial H} \right)_{T.H} \left( \frac{\partial H}{\partial T} \right)_M \right)[/tex]
 
  • #4
This is pretty good. Remember, though, that the temperature derivative is [itex]\partial/\partial T[/itex] (rather than [itex]1/\partial T[/itex], and this may have just been a typo), and that the condition of constant H or M is attached to the derivative only, so there's no [itex]((\partial S/\partial T)_H)_M[/itex], there's only [itex](\partial S/\partial T)_H[/itex], because the M condition vanishes with [itex](\partial T/\partial T)_M[/itex].

Did you notice that your expression for [itex]C_M[/itex] contains [itex]C_H[/itex]? :smile:

Keep going!
 
  • #5
I think I got it, but there are two answers:

[tex]C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H [/tex]

and also:

[tex]C_H-C_M = - T \left( \frac{\partial S}{\partial H} \right)_T \left( \frac{ \partial H}{\partial T} \right)_M [/tex]

Are these both correct? Thanks again for that helpful hint :)

The second part of the question asks obtain an expression for the heat capacity difference in terms of the field (H), temperature (T), and the Curie constant (a).

Using Curie's law, we are given the approximation that:

[tex] \mu_o M/H \approx a/T[/tex]
[tex] M \approx \frac{aH}{T \mu_o}[/tex]

[tex]
C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H = \frac{aTH}{\mu_o} \left( \frac{\partial S}{\partial M} \right)_T
[/tex]

I'm not sure how to change the [tex]\left( \frac{\partial S}{\partial M} \right)_T[/tex] to the variables stated... am I on the right track?
 
  • #6
Pretty much. Your first two expressions are equivalent, which you can show with Maxwell relations. And you can handle [itex](\partial S/\partial M)_T[/itex] with a Maxwell relation. But I'm not following how you differentiated M with respect to T to get your current answer.
 
  • #7
Whoops, this is what i meant:
[tex]

C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H = - \frac{aH}{T\mu_o} \left( \frac{\partial S}{\partial M} \right)_T

[/tex]

The Maxwell relation I got when using the Helmholtz free energy is:

[tex]
\left( \frac{\partial S}{\partial H} \right)_T = \left( \frac{\partial (VM)}{\partial T} \right)_H
[/tex]

Since M is constant, I can't get the relationship with dS/dM?
 
  • #8
There's a Maxwell relation for [itex](\partial S/\partial M)_T[/itex], but it requires a different potential, one that includes dT and dM.
 
  • #9
Here's another shot:

[tex] \left( \frac{\partial S}{\partial M} \right)_T = VM \frac{ \partial ^2 H }{\partial T \partial M} = \frac{V M \mu_o}{a} [/tex]

[tex]
C_H-C_M = - \frac{aH}{T\mu_o} \left( \frac{\partial S}{\partial M} \right)_T = -\frac{HVM}{T}
[/tex]
 
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  • #10
Mapes said:
This is pretty good. Remember, though, that the temperature derivative is [itex]\partial/\partial T[/itex] (rather than [itex]1/\partial T[/itex], and this may have just been a typo), and that the condition of constant H or M is attached to the derivative only, so there's no [itex]((\partial S/\partial T)_H)_M[/itex], there's only [itex](\partial S/\partial T)_H[/itex], because the M condition vanishes with [itex](\partial T/\partial T)_M[/itex].

Did you notice that your expression for [itex]C_M[/itex] contains [itex]C_H[/itex]? :smile:

Keep going!

Also, why wouldn't we use the chain rule to evaluate the derivative? That is...

[tex] H \frac{\partial }{\partial T} \left( \left( \frac{ \partial S}{\partial H} \right)_T \right)_H + \left( \frac{\partial H}{\partial T} \right)_H \left( \frac{ \partial S}{\partial H} \right)_T [/tex]?
 
  • #11
I get MVH/T after trying a couple different ways, so I think you might have a sign error somewhere in your Maxwell relation, but in any case you know the general approach now.

We do always use the chain rule to evaluate the derivative of a product, but differential terms like dH will always be negligible compared to derivatives like [itex](\partial H/\partial T)_M[/itex], so we end up dropping these terms. I'm not sure if this answers your question?
 

1. What is Helmholtz Free Energy?

Helmholtz Free Energy, also known as Helmholtz Function, is a thermodynamic potential that measures the amount of energy available to do useful work in a system at a constant temperature and volume. It is denoted by the symbol F and is defined as F = U - TS, where U is the internal energy, T is the absolute temperature, and S is the entropy of the system.

2. How is Helmholtz Free Energy related to other thermodynamic potentials?

Helmholtz Free Energy is related to other thermodynamic potentials through mathematical equations. It is related to the internal energy (U) through the equation F = U - TS, to the Gibbs Free Energy (G) through the equation F = G - PV, and to the Enthalpy (H) through the equation F = H - TS. These relationships are used to calculate one thermodynamic potential from another.

3. What is the significance of Helmholtz Free Energy in thermodynamics?

Helmholtz Free Energy is significant because it provides a measure of the thermodynamic stability of a system. It tells us whether a system can spontaneously undergo a change in state at a constant temperature and volume. If the Helmholtz Free Energy decreases, the system is thermodynamically stable and spontaneous change can occur. If it increases, the system is unstable and will resist change.

4. How is the C_H-C_M relation used to calculate Helmholtz Free Energy?

The C_H-C_M relation is a mathematical relationship between the specific heat at constant volume (C_V) and the specific heat at constant pressure (C_P) of a substance. It is given by the equation C_H = C_P - TVα, where T is the temperature, V is the volume, and α is the coefficient of thermal expansion. This relation is used to calculate Helmholtz Free Energy through the equation F = U - TS = U - T(C_P - TVα).

5. What experimental techniques are used to determine the C_H-C_M relation?

The C_H-C_M relation can be determined experimentally through various techniques such as calorimetry, differential scanning calorimetry, and thermal analysis. These techniques involve measuring the heat capacity of a substance at constant volume (C_V) and at constant pressure (C_P) at different temperatures. The C_H-C_M relation can then be calculated using the experimental data.

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