# Thermodynamic equation of differentials (and how to work with them)

• FriendlyHamiltonian

#### FriendlyHamiltonian

Disclaimer: I am not a mathematician, I am a physicist.

The thermodynamic identity is usually expressed in the following differential form

$$dU = TdS - PdV + \mu dN,$$

where $U$ , $T$, $S$, $P$, $V$, $\mu$ and $N$ are the internal energy, temperature, entropy, pressure, volume, chemical potential and number of particles of the system respectively. If I am not mistaken, I can act with a vector, say $\frac{\partial}{\partial N}$, to yield

$$\frac{\partial U}{\partial N} = T \frac{\partial S}{\partial N} - P \frac{\partial V}{\partial N} + \mu \implies \mu = \frac{\partial U}{\partial N} - T \frac{\partial S}{\partial N} + P \frac{\partial V}{\partial N}.$$

Consider the following question:

Consider a monoatomic ideal gas that lives at height $z$ above sea level, so each molecule has potential energy $mgz$ in addition to its kinetic energy. Show that the chemical potential $\mu$ is the same as if the gas were at sea level, plus am additional term $mgz$:

$$\mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}\right] + mgz.$$

My attempt was knowing that:

The "ideal monoatomic gas" implies $U = \frac{3}{2}k_bT$ (by equipartition theorem) and the validity of Sackur-Tetrode equation:

$$S=k_bN\ln \left[{\frac {V}{N}}\left({\frac {4\pi m}{3h^{2}}}{\frac {U}{N}}\right)^{3/2}\right]+{\frac {5}{2}},$$

together with the assumption that $V \neq V(N)$. If one uses the above formula for $\mu$ and takes the partial derivatives I yield

$$\mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}-\frac{3}{2}\right] + mgz,$$

which is almost correct except for that $-\frac{3}{2}$, although it still exhibits the problems described below.

I came to the conclusion that I don't know how to manipulate these equations in differential form, am I allowed to do the above "act with $\frac{\partial}{\partial N}$" business? The solution provided by the book is to say, hey hold $U$ and $V$ fixed so that the thermodynamic identity now reads

$$0 = TdS - 0 + \mu dN \implies \mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}}$$

but $U = U(N)$, in particular $U = \frac{3}{2} k_b N T$ I could litterally make all the $N$s in $S$ disappear by substituting $N = \frac{2 U}{3 k_b T}$ and claim that

$$\mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}} = 0,$$

which is ridiculous. I'm really lost with the mathematics behind this type of calculations... Which would be the correct way to proceed?