# Calculating specific heat capacity from entropy

I'm currently struggling with a specific thermodynamics problem. I'm given the entropy of a system (where A is a constant with fitting physical units): $$S(U,V,N)=A(UVN)^{1/3}$$

I'm asked to calculate the specific heat capacity at constant pressure $$C_p$$ and at constant volume $$C_V$$. I know that the two are given by the following equation: $$C_P=T \left(\frac{\partial S}{\partial T}\right)_{N,P}$$ $$C_V=T \left(\frac{\partial S}{\partial T}\right)_{N,V}$$

I've tried to calculate a expression for the entropy in terms of the temperature $$T$$ by calculating $$\left (\frac{\partial S}{\partial U}\right)=\frac{1}{T}=\frac{1}{3}(NV)^{1/3}U^{-2/3}=\frac{1}{3}\frac{S}{U}$$

but I think i'm going in a circle with this approach.

I guess I'm also having problems with the notation $$\left(\frac{\partial S}{\partial T}\right)_{N,P}$$ vs $$\left(\frac{\partial S}{\partial T}\right)_{N,V}$$. I know that this should mean "partial derivative of $$S$$ with respect to $$T$$ with constant $$N$$ and $$P$$ or $$V$$". But I'm not quite sure on how to evaluate that mathematically. For me personally, if I were to evaluate this two partial derivatives, they would be equal to eachother. But this can't be really true because of the relationship $$C_P-C_V=\frac{TV\alpha_P^2}{\kappa_T}$$

It would be awesome if anyone could give me starting point for this kind of problem because I'm kinda stuck with my method right now. Any help would be very much appreciated

## 2 Answers

Try rewriting $$S$$ as a function of $$N$$, $$P$$, $$T$$, and $$V$$ only, using the expressions you know. After that, apply your equations for $$C_P$$ and $$C_V$$. Partial derivative just means you treat every variable other than the one you are differentiating with as a constant.

It might help to review some relevant Calculus. Introductory Thermodyamics is mostly just multivariable Calculus anyway: https://tutorial.math.lamar.edu/classes/calciii/partialderivatives.aspx

• Thanks for taking the time to answer and for the link, I'll look into it! I don't exactly know why I need to write $S$ in terms of $N, P$ and $V$. Doesn't $S$ need to be atleast also a function of $T$ so I can evaluate the partial derivative with respect to $T$? Or am I missing something here? May 2 at 8:12
• Sorry, I meant to include $T$ as well. The point is, you need to eliminate $U$. May 2 at 8:15
• Alright, thanks! So now I computed $$\frac{\partial S}{\partial V}=\frac{P}{T}=\frac{1}{3}(NU)^{1/3}V^{-2/3}$$ and rearranged to get $$U=27\frac{P^3}{T^3}V^2N^{-1}$$ Plugging that back into my orginial equation gives me: $$S(P,V,T) =\frac{PV}{T}$$ So I also eliminated $N$ in the process. Does that seem correct so far? The problem for me is now computing $\left(\frac{\partial S}{\partial T}\right)_{N,P}$ and $\left(\frac{\partial S}{\partial T}\right)_{N,V}$. For me they are both $$\frac{\partial S}{\partial T}=-\frac{PV}{T^2}$$ I guess that isn't correct since $C_P$ and $C_V$ would cancel.. May 2 at 9:17
• In the last differentiation in your comment, it looks like you're holding both $P$ and $V$ constant. Why? Only one of them is held constant in $\left(\frac{\partial S}{\partial T}\right)_{N,P}$ and $\left(\frac{\partial S}{\partial T}\right)_{N,V}$. May 2 at 17:42

Given the expresstion if entropy (Imagine that it is derived from microcanonical ensemble statistically):

$$S(U,V,N)=A\,\left(UVN\right)^{1/3}. \tag{1}$$

And the temperature $$\frac{1}{T} = \frac{A}{3}(NV)^{1/3}U^{-2/3}. \tag{2}$$ or $$U = \sqrt{\frac{A^3 T^3 N V}{27} }. \tag{3}$$

The heat capacity of constant volume is simply $$C_V = \left(\frac{\partial U}{\partial T}\right)_{N, V} = \frac{3}{2} \sqrt{\frac{A^3 T N V}{27} } = \frac{3}{2}\frac{U}{T}. \tag{4}$$

Eq. (4) is the same as derived from your equation using chain-rule: $$C_V=T \left(\frac{\partial S}{\partial T}\right)_{N,V} = T \left(\frac{\partial S}{\partial U}\right)_{N,V} \left(\frac{\partial U}{\partial T}\right)_{N,V}=T \left(\frac{1}{T}\right) \left(\frac{\partial U}{\partial T}\right)_{N,V}= \left(\frac{\partial U}{\partial T}\right)_{N,V}.$$

To calculate $$C_P$$, we first expression the internal energy $$U = U(S,N,V)$$ by inverting Eq.(1) $$U = \frac{S^3}{A^3 V N}. \tag{5}$$

And find the expression of $$P$$ using another Maxwell relation: $$P = -\left(\frac{\partial U}{\partial V}\right)_{S,N}$$.

\begin{align} P = &\,-\left(\frac{\partial U}{\partial V}\right)_{S,N} \\ = & \, \frac{S^3}{A^3 V^2 N} = \, \frac{U}{V}. \tag{6} \end{align}

Differentiate Eq.(6) $$dP = \frac{1}{V} dU - \frac{U}{V^2} dV.$$

Therefore, under condition of constant pressure, $$dP=0$$. We have $$\left(\frac{\partial V}{\partial U}\right)_{P,N} = \frac{V}{U}. \tag{7}$$

Now, we are ready to calculate $$C_P$$. Note that the enthalpy $$H = U + PV = 2PV$$ from Eq.(6). Therefore \begin{align} C_P =& \left(\frac{\partial H}{\partial T}\right)_{P, N} = 2 \left(\frac{\partial PV}{\partial T}\right)_{P, N} = 2P \left(\frac{\partial V}{\partial T}\right)_{P, N} = 2P \left(\frac{\partial U}{\partial T}\right)_{V, N} \left(\frac{\partial V}{\partial U}\right)_{P, N} \\ = & \, 2\,P \, \frac{V}{U}\left(\frac{\partial U}{\partial T}\right)_{V,N} = \, 2\, C_V. \end{align}

In this derivation, we apply the result in Eq.(7) and recall that $$U = PV$$.