- #1
FriendlyHamiltonian
- 1
- 0
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 Ns 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?
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 Ns 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?