Maxwell Relations: Derivations for Enthalpy and Entropy

[onye nkusi]

Maxwell Relations - derivations

Homework Statement

1. Derive the Maxwell Relation based on the enthalpy.
2. Derive the Maxwell Relation based on the entropy.

Homework Equations

H=U+PV
dU=dq+dw
dw=-PdV
dS=dq/T

The Attempt at a Solution

1. I feel like I've gotten this one, but my final may or may not be off.
H=U+PV
dH=dU-d(PV)
dU=dq+dw --- dS=dq/T --- dw=-PdV

dU=TdS-PdV
dH=TdS-PdV+PdV+VdP --- This part I'm unsure about.
dH=TdS+VdP

Finally... (δT/δP)s=-(δV/δS)p

2.
dS=dq/T
TdS=dq
(δT/δq)s=(X/δS) --- I'm absolutely stuck here.

Any input you guys have for either problem, especially the second, is highly appreciated. Thanks.

 

[mark726]

Thank you for your post. Here are the derivations for the Maxwell Relations based on enthalpy and entropy:

1. Derivation based on enthalpy:

Starting with the definition of enthalpy, H = U + PV, we can take the differential of both sides:

dH = dU + PdV + VdP

Since dU = dq + dw, and dw = -PdV, we can substitute these expressions into the above equation:

dH = dq + (-PdV) + PdV + VdP

Simplifying, we get:

dH = dq + VdP

Now, from the definition of entropy, we know that dS = dq/T. Substituting this into the above equation, we get:

dH = TdS + VdP

This is the Maxwell Relation based on enthalpy. To derive the Maxwell Relation based on entropy, we can use this relation and manipulate it to get the desired result.

2. Derivation based on entropy:

Starting with the definition of entropy, dS = dq/T, we can take the differential of both sides:

dS = (δq/δT)p dT + (δq/δp)T dp

Since dS = dq/T, we can substitute this into the above equation:

dq/T = (δq/δT)p dT + (δq/δp)T dp

Rearranging this equation, we get:

(δq/δT)p = T(δS/δT)p

This is the Maxwell Relation based on entropy. We can also manipulate this equation to get the Maxwell Relation based on enthalpy.

I hope this helps. Let me know if you have any further questions or need clarification on any part of the derivation.