Exact Differentials and Partial Derivatives (LearnChemE.com, 5min) This math review puts into context the discussion of exact differentials in Section 6.2 of the textbook using an example related to the volume of a cylinder.

Comprehension Questions:

1. Given that dU = TdS - PdV, what derivative relation comes from setting ∂^{2}U/(∂S∂P) = ∂^{2}U/(∂P∂S)?

2. Given that dA = -SdT - PdV, what derivative relation comes from setting ∂^{2}A/(∂T∂V) = ∂^{2}A/(∂V∂T)?

3. Given that dG = -SdT + VdP, what derivative relation comes from setting ∂^{2}G/(∂T∂P) = ∂^{2}G/(∂P∂T)?

Assembling your derivative toolbox including the triple product rule, (uakron.edu, 13min) Beginning with the fundamental property relation, substitutions lead to Eqns. 6.4-6.7. Differentiating these and equating through exact differentials leads to Eqns. 6.29-6.32 (aka. Maxwell's Relations). Combining Maxwell's Relations with Eqns. 6.4-6.7 leads to Eqns. 6.37-6.41. With these tools in hand, and Eqn. 6.15 (aka. Triple Product Rule), you have all the tools you need to quickly transform any derivative into "expressions involving Cp, Cv, P, V, T, and their derivatives." This capability is fundamental to obtaining expressions for U, H, and S from any given equation of state for any chemical of interest. Four sample derivations are illustrated: (∂U/∂V)_{T}, (∂T/∂S)V, (∂T/∂V)_{S}, (∂S/∂V)A,

Comprehension Questions:

1. Transform the following into "expressions involving Cp, Cv, P, V, T, and their derivatives:" (∂T/∂V)_{S}.

2. Transform the following into "expressions involving Cp, Cv, P, V, T, and their derivatives." Your expression may involve absolute values of S as long as they are not associated with any derivative. (∂T/∂U)_{P}.

Heat Capacity Volume Dependence (uakron.edu, 10min) This example derives how the heat capacity of the gas depends on volume, ie. (∂Cv/∂V)_{T}. It may seem paradoxical that a quantity defined at constant volume can change with respect to volume. The discussion here shows how to solve this puzzle. The sample derivation presented here follows an alternative approach to what is illustrated in Example 6.9 of the textbook.

Comprehension Questions:

1. The van der Waals (vdW) equation of state (EOS) is: P = RT/(V-b) - a/V^{2}. Evaluate (∂Cv/∂V)_{T} for the vdW EOS. 2. The Soave-Redlich-Kwong (SRK) EOS is: P = RT/(V-b) - a/[V(V+b)] where a=[1+K*(1-sqrt(T/Tc))]^{2}. Evaluate (∂Cv/∂V)_{T} for the SRK EOS. 3. Comment on the differences between the results for 1 and 2 above. Do these results change the way you look at the vdW EOS?

## Comments

Lira replied on Permalink

## Derivative Toolbox

Exact Differentials and Partial Derivatives (LearnChemE.com, 5min) This math review puts into context the discussion of exact differentials in Section 6.2 of the textbook using an example related to the volume of a cylinder.

Comprehension Questions:

1. Given that dU = TdS - PdV, what derivative relation comes from setting ∂

^{2}U/(∂S∂P) = ∂^{2}U/(∂P∂S)?2. Given that dA = -SdT - PdV, what derivative relation comes from setting ∂

^{2}A/(∂T∂V) = ∂^{2}A/(∂V∂T)?3. Given that dG = -SdT + VdP, what derivative relation comes from setting ∂

^{2}G/(∂T∂P) = ∂^{2}G/(∂P∂T)?Elliott replied on Permalink

## Derivative Toolbox

Assembling your derivative toolbox including the triple product rule, (uakron.edu, 13min) Beginning with the fundamental property relation, substitutions lead to Eqns. 6.4-6.7. Differentiating these and equating through exact differentials leads to Eqns. 6.29-6.32 (aka. Maxwell's Relations). Combining Maxwell's Relations with Eqns. 6.4-6.7 leads to Eqns. 6.37-6.41. With these tools in hand, and Eqn. 6.15 (aka. Triple Product Rule), you have all the tools you need to quickly transform any derivative into "expressions involving Cp, Cv, P, V, T, and their derivatives." This capability is fundamental to obtaining expressions for U, H, and S from any given equation of state for any chemical of interest.

Foursample derivationsare illustrated: (∂U/∂V), (∂_{T}T/∂S)V, (∂T/∂V), (∂_{S}S/∂V)A,Comprehension Questions:

1. Transform the following into "expressions involving

Cp, Cv, P, V, T, and their derivatives:" (∂T/∂V)._{S}2. Transform the following into "expressions involving

Cp, Cv, P, V, T,and their derivatives." Your expression may involve absolute values ofSas long as they are not associated with any derivative. (∂T/∂U)._{P}Elliott replied on Permalink

## Pressure Dependence of Heat Capacity: (dCp/dP)T

Heat Capacity Pressure Dependence (LearnChemE.com, 7min) This example derives how the heat capacity of the gas depends on pressure, ie. (∂

Cp/∂P)_{T}.Comprehension Questions:

1. Derive the relation for (∂

Cv/∂V)_{T}2. The van der Waals (vdW) equation of state (EOS) is:

P = RT/(V-b) -a/V^{2}. Evaluate (∂Cp/∂P)_{T}for the vdW EOS.Elliott replied on Permalink

## Heat Capacity Volume Dependence:(dCv/dV)T

Heat Capacity Volume Dependence (uakron.edu, 10min) This example derives how the heat capacity of the gas depends on volume, ie. (∂

Cv/∂V)_{T}. It may seem paradoxical that a quantity defined at constant volume can change with respect to volume. The discussion here shows how to solve this puzzle. Thesample derivationpresented here follows an alternative approach to what is illustrated in Example 6.9 of the textbook.Comprehension Questions:

1. The van der Waals (vdW) equation of state (EOS) is:

P = RT/(V-b) -a/V^{2}. Evaluate (∂Cv/∂V)_{T}for the vdW EOS.2. The Soave-Redlich-Kwong (SRK) EOS is:

P = RT/(V-b) -a/[V(V+b)]where

a=[1+K*(1-sqrt(T/Tc))]^{2}. Evaluate (∂Cv/∂V)_{T}for the SRK EOS.3. Comment on the differences between the results for 1 and 2 above. Do these results change the way you look at the vdW EOS?