How to Derive the Equation for Internal Energy in Thermodynamics Homework?

Click For Summary
SUMMARY

The equation for internal energy in thermodynamics is derived as U = -T(∂A/∂T)V, where U represents internal energy, T is temperature, and A is the Helmholtz function. This derivation utilizes the relationships dA = -PdV - SdT and A = U - TS. By recognizing A as a state function and applying partial derivatives, the relationship between the Helmholtz function and internal energy is established. The solution confirms the correctness of the derived equation despite potential issues in the problem statement.

PREREQUISITES
  • Understanding of thermodynamic functions, specifically Helmholtz function.
  • Familiarity with partial derivatives in the context of state functions.
  • Knowledge of thermodynamic equations, particularly dA = -PdV - SdT.
  • Basic concepts of internal energy and its relation to temperature and entropy.
NEXT STEPS
  • Study the derivation of the Helmholtz free energy and its applications in thermodynamics.
  • Explore the implications of state functions in thermodynamic processes.
  • Learn about the relationships between internal energy, temperature, and entropy in greater detail.
  • Investigate the role of partial derivatives in thermodynamic equations and their physical significance.
USEFUL FOR

Students studying thermodynamics, educators teaching thermodynamic principles, and researchers focusing on energy systems and thermodynamic modeling.

arpon
Messages
234
Reaction score
16

Homework Statement


Derive the equation
##U=-T(\frac{\partial A}{\partial T})_V##
where ##U## is the internal energy, ##T## is the temperature, ##A## is the Helmholtz function.
Reference: Heat and Thermodynamics, Zemansky, Dittman, Page 272, Problem 10.4 (a)

Homework Equations


##dA=-PdV-SdT## ... (i) [##S## is the entropy]
##A=U-TS## ... (ii)

The Attempt at a Solution


As ##A(V,T)## is a state function,
##dA = (\frac{\partial A}{\partial V})_T dV + (\frac{\partial A}{\partial T})_V dT## ... (iii)
Comparing (i) and (iii),
##(\frac{\partial A}{\partial V})_T = -P## ... (iv)
and ##(\frac{\partial A}{\partial T})_V = -S## ... (v)
Using (ii) and (v),
##U = A +TS = A- T(\frac{\partial A}{\partial T})_V##
Any help would be appreciated.
 
Last edited:
Physics news on Phys.org
I think your solution is correct while the problem as stated is wrong.
 
  • Like
Likes   Reactions: arpon and Chestermiller

Similar threads

Replies
30
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
2
Views
7K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
4
Views
1K
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K