Please help me to derive that ${(\frac{\partial S}{\partial T})}_{P} = \frac{C_{P}}{T}$ $((delS)/(delT))_P = C_P/T$ and ${(\frac{\partial S}{\partial T})}_{V} = \frac{C_{V}}{T}$ $((delS)/(delT))_V = C_V/T$ ?

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Please help me to derive that ${(\frac{\partial S}{\partial T})}_{P} = \frac{C_{P}}{T}$ $((delS)/(delT))_P = C_P/T$ and ${(\frac{\partial S}{\partial T})}_{V} = \frac{C_{V}}{T}$ $((delS)/(delT))_V = C_V/T$ ?

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Answer 1 · 2017-01-25T06:26:00

Okay, so I will assume you know or can figure out the Maxwell relations. Those are a common starting point. Furthermore, you have by definition that $C_{P} = {(\frac{\partial H}{\partial T})}_{P}$ $C_P = ((delH)/(delT))_P$ and $C_{V} = {(\frac{\partial U}{\partial T})}_{V}$ $C_V = ((delU)/(delT))_V$ , so you do not have to derive those.

Recall that $G = H - T S$ $G = H - TS$ . Then, the derivative is:

$d G = d H - d (T S)$ $dG = dH - d(TS)$

$= d H - S d T - T d S$ $= dH - SdT - TdS$
(use product rule)

Now, for the first derivation, you can divide through by $\partial T$ $delT$ at a constant $P$ $P$ to generate two derivatives that can be figured out, and the third one is what you're looking for:

$((delG)/(delT))_P = ((delH)/(delT))_P - cancel(S((delT)/(delT))_P)^(1) - T((delS)/(delT))_P$

$" "" "bb((1))$

where the third term goes to $1$ because $T$ does not change with respect to $T$ ; it is one-to-one with itself.

By definition, $((delH)/(delT))_P = C_P$ . $" "" "bb((2))$

Next, recall that for the natural variables $T$ and $P$ correspond to the Gibbs' free energy, so that the Maxwell relation is:

$dG = -SdT + VdP$ $" "" "" "bb((3))$

We do not yet know what $((delG)/(delT))_P$ is, but using the Maxwell relation, we get:

$((delG)/(delT))_P = -S$ $" "" "" "bb((4))$

Therefore. plugging $bb((4))$ and $bb((2))$ into $bb((1))$ :

$cancel(-S) = C_P - cancel(S) - T((delS)/(delT))_P$

It follows that if $C_P = T((delS)/(delT))_P$ , it must be that $color(blue)(((delS)/(delT))_P = C_P/T)$ .

A similar process follows for deriving $C_V/T = ((delS)/(delT))_V$ . Using:

$A = U - TS$ ,

take the derivative to get:

$dA = dU - SdT - TdS$

Almost like before, divide by $delT$ at a constant $V$ instead of $P$ to generate two derivatives that can be figured out, and the third one is what you're looking for:

$((delA)/(delT))_V = ((delU)/(delT))_V - Scancel(((delT)/(delT))_V)^(1) - T((delS)/(delT))_V$

We know by definition that $((delU)/(delT))_V = C_V$ , $((delT)/(delT))_V = 1$ , and that $A$ is a function of the natural variables $T$ and $V$ .

So from the Maxwell relation:

$dA = -SdT - PdV$

we have that:

$((delA)/(delT))_V = -S$ ,

so plugging back into the main equation, we get:

$cancel(-S) = C_V - cancel(S) - T((delS)/(delT))_V$

Since $C_V = T((delS)/(delT))_V$ , it follows that $color(blue)(((delS)/(delT))_V = C_V/T)$ .

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Please help me to derive that ${(\frac{\partial S}{\partial T})}_{P} = \frac{C_{P}}{T}$ $((delS)/(delT))_P = C_P/T$ and ${(\frac{\partial S}{\partial T})}_{V} = \frac{C_{V}}{T}$ $((delS)/(delT))_V = C_V/T$ ?

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Please help me to derive that ((delS)/(delT))_P = C_P/T(∂S∂T)P=CPT((delS)/(delT))_P = C_P/T and ((delS)/(delT))_V = C_V/T(∂S∂T)V=CVT((delS)/(delT))_V = C_V/T?

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Please help me to derive that ${(\frac{\partial S}{\partial T})}_{P} = \frac{C_{P}}{T}$ $((delS)/(delT))_P = C_P/T$ and ${(\frac{\partial S}{\partial T})}_{V} = \frac{C_{V}}{T}$ $((delS)/(delT))_V = C_V/T$ ?