Equation of state: Ideal gas example for energy, entropy, temperature

Question

I'm trying to apply the definition of temperature $$\frac{\partial{U}}{ \partial{S}} \equiv T$$ to a simple example: an ideal gas expanding adiabatically in a piston. The internal energy is decreasing, because the piston is moving outwards and the KE drops during piston collisions and the entropy is increasing. Is there a sign problem? T should be negative by this definition.

Edit: I realized that the partial derivative requires that the other variables (V,N) must be fixed. I'm not just changing S, I'm changing V in my example.

Is there a simple ideal gas example of this equation?

Answer 1

State variables and partial derivatives in thermodynamics. In order to use partial derivatives, you

• first need the expression of a state variable as a function of other two state variables;
• then evaluate the partial derivatives you like, keeping in mind the constant independent variable.

!!!Working with multi-variable functions, and with change of variables is 90% of algebra and calculus needed to derive foundations of all the classical thermodynamics!!! (this hint will never be BOLD ENOUGH)

As an example, using the first principle of thermodynamics with $U(V,S)$ or $S(V,U)$

$$dU = - P dV + T dS \qquad \text{or} \qquad dS = \frac{1}{T} dU + \frac{P}{T} dV$$

where the following holds,

$$T = \left( \frac{\partial U}{\partial S} \right)_V \qquad , \qquad \frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_V$$ $$P = -\left( \frac{\partial U}{\partial V} \right)_S \qquad , \qquad \frac{P}{T} = \left(\frac{\partial S}{\partial V}\right)_U$$

being $dU(V,S)$ and $dS(V,U)$ perfect differentials, or differentials of state functions.

Temperature as partial derivative, for a perfect gas. Thus, in your example using perfect gases - independently from the transformation you're performing -, you first need to write internal energy as a function of $T$ and $S$ and then evaluate the partial derivative. Equation of state and internal energy for a perfect gas read,

$$P V = n R T \qquad , \qquad U = C_V T \ ,$$

while it's possible to write change in entropy as

$$\begin{aligned} S - S_0 & = C_V \ln\left( \frac{T}{T_0} \right) + n R \ln\left( \frac{V}{V_0} \right) = \\ & = C_V \ln\left( \frac{U}{U_0} \right) + n R \ln\left( \frac{V}{V_0} \right) \ . \end{aligned}$$

Now that you have an expression of the entropy as a function of internal energy and volume, $S(U,V)$, you can perform the partial derivative remembering that $\frac{d}{dx} \ln \left( \frac{x}{x_0} \right) = \frac{1}{x}$, to get

$$\left(\frac{\partial S}{\partial U}\right)_V = \frac{C_V}{U} = \frac{1}{T} \ ,$$

i.e. the desired result. This result is independent from the transformation, as you would expect being "a definition of temperature".

Your example. In an adiabatic expansion, without dissipation, $S$ is constant. In order to describe the state of the system during the thermodynamic transformation, you need the value of another thermodynamic state variable, $X$ (pressure, temperature,... you're free to choose). Anyway you can't expect to get any useful result from differentiation of the function $U(X,S)$, since $V$ is varying (and this contradicts the definition of temperature) and $S$ is constant,

$$T \neq \left( \frac{\partial S}{\partial U} \right)_x = 0 \qquad \text{for every $X$}$$

A working example. In order to use the definition of $T$ having all the required physical quantities and thermodynamic transformation, you need a system that performs a transformation at constant volume: as an example, you can provide heat flux to a gas in a constant volume rigid box.

In this case, $V$ is const, and $dU(\overline{V},S) = T(S) dS$. For constant volume, energy can be written a single variable function of $S$, whose derivative w.r.t. $S$ (at constant $V$) is the temperature.

Answer 2

Yes, there is a simple ideal-gas example of $\left(\frac{\partial U}{\partial S}\right)_{V,N}=T$: heating it. From the First Law, if the volume remains constant, the energy change $\partial U$ corresponds to infinitesimal heating $\delta Q$, and we find that the corresponding entropy change is $dS=\frac{\delta Q}{T}$.

This leads to many useful conclusions. One is that the heating can’t be occurring spontaneously from an adjacent cooler ideal gas at temperature $T^\prime<T$, as that gas would be losing more entropy than the first gas gains, and total entropy destruction is prohibited by the Second Law.