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

In summary, the thermodynamic identity is usually expressed in the following differential form. 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. The chemical potential is the same as if the gas were at sea level, plus am additional term mgz.
  • #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 [itex] U [/itex] , [itex] T [/itex], [itex] S [/itex], [itex] P [/itex], [itex] V [/itex], [itex] \mu [/itex] and [itex] N [/itex] 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 [itex]\frac{\partial}{\partial N}[/itex], 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 [itex]z[/itex] above sea level, so each molecule has potential energy [itex]mgz[/itex] in addition to its kinetic energy. Show that the chemical potential [itex]\mu[/itex] is the same as if the gas were at sea level, plus am additional term [itex]mgz[/itex]:

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

The "ideal monoatomic gas" implies [itex]U = \frac{3}{2}k_bT[/itex] (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 [itex]V \neq V(N)[/itex]. If one uses the above formula for [itex]\mu[/itex] 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 [itex]-\frac{3}{2}[/itex], 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 [itex]\frac{\partial}{\partial N}[/itex]" business? The solution provided by the book is to say, hey hold [itex]U[/itex] and [itex]V[/itex] 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 [itex]U = U(N)[/itex], in particular [itex]U = \frac{3}{2} k_b N T[/itex] I could litterally make all the [itex]N[/itex]s in [itex]S[/itex] disappear by substituting [itex]N = \frac{2 U}{3 k_b T}[/itex] 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?
 
Science news on Phys.org
  • #2
Here gravity or gravitational energy is considered for chemical potential. It is not considered for internal energy and Sackur-Tetrode equation. Why don't you treat them in a same way ?
 

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

What is the thermodynamic equation of differentials?

The thermodynamic equation of differentials is a mathematical expression that describes the relationship between the changes in thermodynamic properties of a system. It is based on the first and second laws of thermodynamics and is commonly used in thermodynamic calculations and analysis.

How do I work with thermodynamic equations of differentials?

To work with thermodynamic equations of differentials, you will need to have a basic understanding of calculus and thermodynamics. This includes knowledge of differential equations, partial derivatives, and thermodynamic properties such as temperature, pressure, and volume. It is also helpful to have a good understanding of the first and second laws of thermodynamics and how they relate to the thermodynamic equation of differentials.

What are the key components of the thermodynamic equation of differentials?

The key components of the thermodynamic equation of differentials are the thermodynamic properties, such as temperature, pressure, and volume, and the differential terms, such as dT, dP, and dV. These components represent the changes in the thermodynamic properties of a system and are used to calculate the changes in energy, work, and heat.

How is the thermodynamic equation of differentials used in thermodynamic calculations?

The thermodynamic equation of differentials is used in thermodynamic calculations to determine the changes in energy, work, and heat in a system. It is also used to calculate the efficiency of thermodynamic processes and to analyze the behavior of different thermodynamic systems.

Are there any limitations to the thermodynamic equation of differentials?

While the thermodynamic equation of differentials is a powerful tool in thermodynamics, it does have some limitations. It assumes that the system is in a state of equilibrium, and it does not take into account any irreversible processes. It is also limited in its application to ideal gases and cannot accurately describe the behavior of real gases.

Similar threads

Replies
6
Views
1K
Replies
2
Views
569
Replies
19
Views
1K
  • Thermodynamics
Replies
4
Views
1K
Replies
23
Views
1K
Replies
1
Views
879
Replies
3
Views
730
Replies
1
Views
979
  • Thermodynamics
Replies
2
Views
853
Replies
7
Views
813
Back
Top