| New Reply |
Fundamental thermodynamic relation confusion. |
Share Thread | Thread Tools |
| May9-12, 08:26 AM | #1 |
|
|
Fundamental thermodynamic relation confusion.
dE =
We have for an reversible process, dE = TdS - PdV So this relation is for all changes (irreversible or reversible) since dS and dV are state functions. What doesn't make sense to me is the next part when Helmholtz free energy is defined; F = E - TS Then dF = -PdV - SdT. Another relation for all changes. I'm told and shown that for a system at constant temperature, then ΔW ≥ ΔF, with equality for reversible processes. However how is this equality true ONLY for reversible processes? If it's at constant temperature, then dT = 0. So dF = -PdV = dW no matter what process is? |
| May9-12, 09:16 AM | #2 |
|
|
But -PdV = dW only for reversible processes (as you said yourself in the second line of your post).
|
| May9-12, 09:39 AM | #3 |
|
|
|
| May9-12, 09:47 AM | #4 |
|
|
Fundamental thermodynamic relation confusion.
I'm not sure if I understand your problem, but it seems as if you're under the impression that dE = Q + W = TdS - PdV implies that Q = TdS and W - PdV? Is that correct? If so, that is where your mistake is.
|
| May9-12, 10:08 AM | #5 |
|
|
|
| May9-12, 10:16 AM | #6 |
|
|
The idea is as follows: take any two states of a system. No matter what process you use to go from the first state to the second state, dE, dS and dV are always the same, by definition of state variables (i.e. they only depend on the state, not on the process). So TdS - PdV is a quantity that only depends on the initial and the final state, not on the process that connects these two states. And since we know that dE = TdS - PdV is true for at least one process (i.e. the reversible one), and knowing that this equality cannot depend on the process, we know that it most be true for all the processes.
Does that clarify matters? |
| May9-12, 01:13 PM | #7 |
|
|
Deep down what confused me is that my notes say this; "For an infinitesimal reversible process; dF = -PdV - SdT" Although this is true, it's actually for all processes, not just reversible ones. My notes saying "for an infinitesimal reversible process" made me think it was suggesting that this relationship only holds for reversible processes and not irreversible ones. |
| May9-12, 05:27 PM | #8 |
|
|
I think what you're saying is correct yes.
As a caveat, do note that I'm still somewhat confused myself about some of the details related to this discussion (cf. a thread I started in this same subforum inspired by yours). That being said, I'm fairly sure the answers I gave to your questions were correct, and anyway you have your own brain that can judge the arguments accordingly :) |
| New Reply |
| Thread Tools | |
Similar Threads for: Fundamental thermodynamic relation confusion.
|
||||
| Thread | Forum | Replies | ||
| Fundamental Equation Thermodynamic Postulates | Advanced Physics Homework | 0 | ||
| Thermodynamic property relation derivation | Advanced Physics Homework | 0 | ||
| Thermodynamic Relation | Engineering, Comp Sci, & Technology Homework | 2 | ||
| The fundamental thermodynamic relation | Classical Physics | 8 | ||
| Confusion over Thermodynamic process (heat and termperature) | Mechanical Engineering | 3 | ||