Constant Pressure Specific Heat in terms of Entropy and Enthelpy

[PhDeezNutz]

Homework Statement

Per my Statistical Mechanics Book (Pathria and Beale Third Edition) formula 1.3.18 reads

$C_p = T \left( \frac{\partial S}{\partial T}\right)_{N,P} = \left( \frac{\partial \left( E + PV\right)}{\partial T}\right)_{N,P} = \left( \frac{\partial H}{\partial T}\right)_{N,P}$

I do not see how

$T \left( \frac{\partial S}{\partial T}\right)_{N,P}$

(The definition that I provisionally accept) is equal to

$\left( \frac{\partial H}{\partial T}\right)_{N,P}$

When I do it I get an extra term $S$ per the product rule (I'll show my work shortly)

Relevant Equations

Helmholtz Free Energy

$A = E - TS$

Gibbs Free Energy

$G = A + PV = E - TS + PV = \mu N$

Enthalpy

$H = E + PV = G + TS$

If $N$ is constant (per the partial derivatives definitions/ the subscripts after the derivatives) then $G$ is constant

$H - TS = constant$

Taking the derivative of both sides with respect to $T$ while holding $N,P$ constant we get the following with the use of the product rule

$\left( \frac{\partial H}{\partial T}\right)_{N,P} - T \left( \frac{\partial S}{\partial T}\right)_{N,P} - S = 0$

Which then implies

$\left( \frac{\partial H}{\partial T}\right)_{N,P} = T \left( \frac{\partial S}{\partial T}\right)_{N,P} + S$

The above answer is "right" with the exception of the additive factor S but I don't see how to make it disappear.

I've been thinking about this for a day or so and I can't seem to make any progress. So any help is appreciated in advance.

 

[TSny]

The chemical potential $\mu$ is generally a function of $T, P,$ and $N$. So, $\mu N$ cannot be treated as a constant when taking the partial with respect to $T$ at constant $N$ and $P$.

 

[PhDeezNutz]

The chemical potential $\mu$ is generally a function of $T, P,$ and $N$. So, $\mu N$ cannot be treated as a constant when taking the partial with respect to $T$ at constant $N$ and $P$.

Good point. Upon reading your suggestion I decided to take a different approach because I don't know what $\left( \frac{\partial \mu}{\partial T}\right)_{N,P}$ represents.

Lord forgive me for what I'm about to do. I'm going to egregiously abuse notation and math but in the end I get what I want so there must be something correct about it. Let me know what you think if and when you get the chance.

Here we go

$dQ = TdS = c_p dT \Rightarrow c_p = T \left( \frac{\partial S}{\partial T}\right)_{N,P}$Now to derive an alternate expression for $c_p$ and then equate them.

Recall from the thermodynamic identity

$dE = TdS - PdV + \mu dN \Rightarrow TdS = dE + PdV - \mu dN$

since $dN = 0$ per assumption

$TdS = dE + PdV = c_p dT$

Therefore

$c_p = \left(\frac{\partial E}{\partial T}\right)_{N,P} + P \left( \frac{\partial V}{\partial T} \right)_{N,P}$

so we have

$c_p = \left(\frac{\partial H }{\partial T } \right)_{N,P}$

The "egregious" part is me interchanging partial and full derivatives.

 

[TSny]

That all looks good to me.

Regarding the "egregious" part. You wrote

$dQ = TdS = c_p dT$.

The first equality $dQ = TdS$ assumes that $N$ is constant. The equality $dQ = c_p dT$ assumes $P$ is constant. So, the relation $TdS = c_P dT$ assumes $N$ and $P$ are constant. When you rearrange this as a deritvative $c_P = T\frac{dS}{dT}$, the derivative is being done at constant $P$ and $N$. So, it is actually a partial derivative: $c_P = T \left( \frac{\partial S}{\partial T} \right)_{P,N}$

 

[PhDeezNutz]

That all looks good to me.

Regarding the "egregious" part. You wrote

$dQ = TdS = c_p dT$.

The first equality $dQ = TdS$ assumes that $N$ is constant. The equality $dQ = c_p dT$ assumes $P$ is constant. So, the relation $TdS = c_P dT$ assumes $N$ and $P$ are constant. When you rearrange this as a deritvative $c_P = T\frac{dS}{dT}$, the derivative is being done at constant $P$ and $N$. So, it is actually a partial derivative: $c_P = T \left( \frac{\partial S}{\partial T} \right)_{P,N}$

That makes perfect sense. I've always found the ways partial derivatives are manipulated in thermo/statistical mechanics to be somewhat mystifying and your post clears a lot of it up.

Thank you very much.

 

[Chestermiller]

Isn't $dH=TdS+VdP$?

 

[PhDeezNutz]

Isn't $dH=TdS+VdP$?

From my undergrad thermo class I recall enthalpy being the amount of energy to create a system in a constant pressure atmosphere

so we must

1) account for the energy of the system

2) account for the work needed to "push the atmosphere out of the way" (in a constant pressure atmosphere)

so I think the differential is on V instead of P.

 

[Chestermiller]

From my undergrad thermo class I recall enthalpy being the amount of energy to create a system in a constant pressure atmosphere

so we must

1) account for the energy of the system

2) account for the work needed to "push the atmosphere out of the way" (in a constant pressure atmosphere)

so I think the differential is on V instead of P.

Well, I'm sad to say you think wrong.

The definition of enthalpy is $$H\equiv U+PV$$and the fundamental equation relating differential variations in U, S, and V is: $$dU=TdS-PdV$$So, $$dH=dU+PdV+VdP=(TdS-PdV)+PdV+VdP=TdS+VdP$$

This is something fundamental that should have been covered in your undergrad thermo class.

Please don't try to correct other members unless you are sure of what you are saying.

 

[PhDeezNutz]

Well, I'm sad to say you think wrong.

The definition of enthalpy is $$H\equiv U+PV$$and the fundamental equation relating differential variations in U, S, and V is: $$dU=TdS-PdV$$So, $$dH=dU+PdV+VdP=(TdS-PdV)+PdV+VdP=TdS+VdP$$

This is something fundamental that should have been covered in your undergrad thermo class.

Please don't try to correct other members unless you are sure of what you are saying.

Apologies I didn't mean to offend. I should have prefaced my post with "If I recall correctly but my memory is fuzzy". I see that you are a valuable member on these forums and I didn't mean to undermine you.

 

[PhDeezNutz]

@Chestermiller I was thinking about the definition of $dH$ the other day and although I saw that you were right (and I was wrong) I think I have more insight now. I re-read the parts of my undergrad book and saw that they restricted themselves to constant pressure systems (because they are common in nature?) and then concluded $dH=dU+PdV$. But nowhere did they say it had to be a constant pressure environment.

I mistakenly thought this meant there was no differential on $P$ but using the fundamental relation you put forth for $dU$ I see that there is.

If I may ask a question;

In a constant pressure environment is the change in enthalpy equal to the heat added $Tds$?

 

[PhDeezNutz]

Apparently (according to my book)

$\Delta H = Q + W_{other}$ for constant $P$. I don't quite understand that one.

It must be because #TdS# is not equal to heat?

 

[Chestermiller]

The equation dH=TdS+VdP refers to the change in the thermodynamic equilibrium function H(S,P) for a pure substance or a mixture of constant chemical composition between two closely neighboring, differentially separated, thermodynamic states, one at coordinates (S,P) and the other at coordinates (S+dS, P+dP). This relationship is independent of any physical process that took the system from state 1 to state 2, no matter how complex, convoluted, or irreversible the process was, provided only, that in the end, they are again closely neighboring. So if dP between these two states is zero, it follows that dH=TdS.

On the other hand, the equation $\Delta H=Q+W_{other}$ applies to a finite change resulting from a process in which the system is in contact externally with a constant force per unit area over the entire process, including in its initial and final states. If the process is also reversible, then $\Delta S=\int{\frac{dQ}{T}}$. If the temperature is changing during the process, this, of course, does not mean that $Q=\int{TdS}$.