Partial derivatives (Maxwell relations) in thermodynamics

[AxiomOfChoice]

My professor did this in lecture, and I can't figure out his logic. Can someone fill in the gaps?

He went from:

$$dS = \left( \frac{\partial S}{\partial P} \right)_T dP + \left( \frac{\partial S}{\partial T} \right)_P dT$$

(which I totally understand; it just follows from the fact that $S$ is an exact differential) to the following:

$$\left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T}\right)_V + \left( \frac{\partial S}{\partial T} \right)_P$$

Where the heck does THAT come from? Anyone have any ideas?

 

[gabbagabbahey]

My professor did this in lecture, and I can't figure out his logic. Can someone fill in the gaps?

He went from:

$$dS = \left( \frac{\partial S}{\partial P} \right)_T dP + \left( \frac{\partial S}{\partial T} \right)_P dT$$

(which I totally understand; it just follows from the fact that $S$ is an exact differential) to the following:

$$\left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T}\right)_V + \left( \frac{\partial S}{\partial T} \right)_P$$

Where the heck does THAT come from? Anyone have any ideas?

The first equation tells you that $S$ can be written as a function of $T$ and $P$ only... i.e. $S=S(P,T)$...the second equation is just a straight forward application of the chain rule to $S(P,T)$

 

[AxiomOfChoice]

The first equation tells you that $S$ can be written as a function of $T$ and $P$ only... i.e. $S=S(P,T)$...the second equation is just a straight forward application of the chain rule to $S(P,T)$

Thanks for your response, but I'm afraid I don't understand. How can a "straightforward application of the chain rule to $S(P,T)$" involve the variable $V$? I would think such an application could only contain the variables $S, T, P$.

 

[gabbagabbahey]

Thanks for your response, but I'm afraid I don't understand. How can a "straightforward application of the chain rule to $S(P,T)$" involve the variable $V$? I would think such an application could only contain the variables $S, T, P$.

Both $T$ and $P$ may have some dependence on $V$.

$$\left(\frac{\partial S}{\partial T}\right)_V$$

simply means the partial derivative of $S$, with repect to $T$, taken at constant $V$. Whenever you calculate a partial derivative of a multivariable function, you're always holding at least one variable constant.

 

[George Jones]

Whenever you calculate a partial derivative of a multivariable function, you're always holding at least one variable constant.

Actually, all the other variables are constant.

My professor did this in lecture, and I can't figure out his logic. Can someone fill in the gaps?

He went from:

$$dS = \left( \frac{\partial S}{\partial P} \right)_T dP + \left( \frac{\partial S}{\partial T} \right)_P dT$$

(which I totally understand; it just follows from the fact that $S$ is an exact differential) to the following:

$$\left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T}\right)_V + \left( \frac{\partial S}{\partial T} \right)_P$$

Where the heck does THAT come from? Anyone have any ideas?

When it come to the chain rule, often physicists are very sloppy. Start with $S=S\left(P,T\right)$ and suppose that pressure is a function of volume and temperature. Define

$$\tilde{S} \left(V,T\right) = S \left( P\left(V,T\right) , T \right).$$

Now, use the chain rule to calculate $$\partial \tilde{S} / \partial T$$.

Although $S$ and $$\tilde{S}$$ are very related, they actually are different functions.

 

[jambaugh]

Actually, all the other variables are constant.When it come to the chain rule, often physicists are very sloppy. Start with $S=S\left(P,T\right)$ and suppose that pressure is a function of volume and temperature. Define

$$\tilde{S} \left(V,T\right) = S \left( P\left(V,T\right) , T \right).$$

Now, use the chain rule to calculate $$\partial \tilde{S} / \partial T$$.

Although $S$ and $$\tilde{S}$$ are very related, they actually are different functions.

You beat me to it. But let me add. The real problem (sloppiness) is that we confuse the variables with the functions e.g. x = x(t). We should really begin with entropy as a function of T, P,N and V.
S = s(T,P,N,V) so
$$dS = \frac{\partial s}{\partial P} dP + \frac{\partial s}{\partial T} dT + \frac{\partial s}{\partial V}dV + \frac{\partial s}{\partial N}dN$$
But we also have a constraint on T,P, N, and V via the physical gas law. This we can express implicitly f(P,V,N,T)=0, or solve for any of the four quantities as a function of the other three.

$$P = p(V,T,N)$$
$$V = v(P,T,N)$$
$$T = t(P,V,N)$$
$$N = n(P,V,T)$$

Differentiate each of these to relate their partials to f's.

We then may impose a condition such as constancy of volume or temperature, etc.

It may be rather lengthy but I think worthwhile to go through the long derivation then practice converting to the "sloppy" form used.

[EDIT]: Actually "we begin with" the gas law and that is incorporated into the entropy function so I got overly inclusive with the variables.

 

[nj79]

Maybe he used Maxwell relations?
Its a difficult to calculate.
But you can find loads of ways to find similar relations :
G(T,P)