Exercises for the thermodynamic network

Objective

Learn how to derive relations between thermodynamic potentials and get some interesting relationships

Introduction

In the previous section we looked at the thermodynamic network. We now wish to expand on this. First we will look at a procedure for deriving derivatives of thermodynamic potentials, and then (as an exercise) derive some of these derivatives.

Derivatives of thermodynamic potentials

To derive derivatives which involve thermodynamic potentials we can use the following procedures:

1. Move the potential to the numerator. Then use the corresponding form of the fundamental equation to eliminate the potential.
2. Move the entropy to the numerator. If a Maxwell equation can be used to eliminate S then use it. If not divide both numerator and denominator by [math]\partial T[/math]. The numerator can then be written in terms of heat capacities.
3. Move the volume to the numerator. The remaining derivatives can be expressed in terms of α and κ
4. Elimate C_V by the equation [math]C_{V}=C_{P}-\frac{TV/\alpha^{2}}{\kappa_{T}}[/math]

Exercises

In order to better understand the thermodynamic network it is useful to derive some of the derivatives of thermodynamic potentials.

Instructions: For each of the following derivatives involving thermodynamic potentials, simplify them in terms of P, V, T, S, α, κ, and C_P

Hints:

1. You will need to use the fundamental equations and Maxwell equations. They can be found in this table. Also review the derivatives from the previous section.
2. Do not forget the rules for manipulating derivatives.

[math]\left (\frac{\partial U}{\partial T}\right )_P[/math]

[math]\left (\frac{\partial U}{\partial P}\right )_T[/math]

[math]\left (\frac{\partial H}{\partial T}\right )_P[/math]

[math]\left (\frac{\partial H}{\partial P}\right )_T[/math]

[math]\left (\frac{\partial H}{\partial V}\right )_T[/math]

[math]\left (\frac{\partial A}{\partial T}\right )_P[/math]

[math]\left (\frac{\partial A}{\partial P}\right )_T[/math]

[math]\left (\frac{\partial A}{\partial P}\right )_S[/math]

[math]\left (\frac{\partial G}{\partial T}\right )_P[/math]

[math]\left (\frac{\partial G}{\partial P}\right )_T[/math]

[math]\left (\frac{\partial G}{\partial P}\right )_S[/math]

[math]\left (\frac{\partial G}{\partial T}\right )_V[/math]

[math]\kappa_S=-\frac{1}{V}\left (\frac{\partial V}{\partial P}\right )_S[/math]

[math]\mu_S=\left (\frac{\partial T}{\partial P}\right )_S[/math]

[math]\mu_H=\left (\frac{\partial T}{\partial P}\right )_H[/math]

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Notes

1. As an example of the use of thermodynamic potentials, say we want to find the change in internal energy, i. e. ΔU. Remember we cannot directly measure it.

Using the total differential for U gives [math]dU=\left(\frac{\partial U}{\partial P}\right)_{T}dP+\left(\frac{\partial U}{\partial T}\right)_{P}dT[/math]

We can now substitute the results of the exercise for the two partial derivatives and then use PVT and heat capacity data to calculate ΔU.

2. [math]\kappa_S[/math] is called the isentropic compressibility. Isentropic implies an adiabatic process.

3. [math]\mu_H[/math] is called the Joule-Thompson coefficient. The H subscript may seem odd, but a constant enthalpy process occurs when a gas goes through a constriction such as a partially opened valve or a porous plug. One type of this is an expansion valve, which is found in all refrigerators and household air conditioners.