Partial derivatives of enthelpy and Maxwell relations

Like Tony Stark
Messages
182
Reaction score
6
Homework Statement
Write the following partial derivatives in terms of heat capacity (##c##), compressibility (##\kappa##) and coefficient of thermal expansion (##\alpha##).

##(\frac{\partial H}{\partial V})_U##
##(\frac{\partial H}{\partial V})_S##

With ##H##: enthalpy, ##U##: internal energy, ##V##: volume, ##S##: entropy and the subscript denotes a magnitud held constant in differentiation.
Relevant Equations
##dH=dU+Pdv+vdP##
##dU=TdS-PdV##
I've attached images showing my progress. I have used Maxwell relations and the definitions of ##\alpha##, ##\kappa## and ##c##, but I don't know how to continue. Can you help me?
 

Attachments

  • 20210419_215620.jpg
    20210419_215620.jpg
    59.5 KB · Views: 189
Physics news on Phys.org
Your picture is unreadable.
 
You have that ##H = U + PV##. And you also have that ##dU = T dS - P dV##. So when ##U## is constant, you have: ##dH = d(PV) = PdV + V dP##. And you also have when ##U## is constant, ##TdS - PdV = 0##. So we conclude:

##\frac{\partial H}{\partial V}|_U = P + V \frac{\partial P}{\partial V}|_U##

Now, if you think of ##P## as a function of ##S## and ##V##, then

##\frac{\partial P}{\partial V}|_U = \frac{\partial P}{\partial V}|_S + \frac{\partial P}{\partial S}|_V \frac{\partial S}{\partial V}|_U##

At this point, you can use ##TdS - PdV = 0## (when ##dU = 0##) to rewrite ##\frac{\partial S}{\partial V}|_U##, and you can use one of the Maxwell relations to rewrite ##\frac{\partial P}{\partial S}|_V##.
 
  • Like
Likes   Reactions: Like Tony Stark and etotheipi

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
2K