# Enthalpy and work(at constant temperature) are state function

1. ### Outrageous

375
Enthalpy is a state function because it only depends on the initial and final state.
Work is a path function because its value depend on the path.
So can I say work done on a system at constant pressure is a state function ?
Because the path for the work here is always the same, which is under the same pressure??

Thank you

Last edited: Dec 26, 2012
2. ### DrDu

4,420
You have to be careful here. Even if the equilibrium states of a system depend on only one work variable, e.g. V, there are usually an infinity of possibilities to do irreversible work on the system, e.g. by stirring it.
Work becomes a state function for adiabatic processes where it defines the change of internal energy U.

3. ### Outrageous

375
Do you mean that work done in a reversible process and at constant pressure , can be considered as a state function?

By doing experiment, we can prove that in adiabatic process, any work paths between two states will have same adiabatic work. That experiment defines the change of internal energy. So can I say work in adiabatic process is a state function?

4. ### 256bits

State functions depend only upon the state of the system.
Path functions depend upon the path taken from state 1 to 2.

Can one define the work W1 at state 1?

DrDu alludes to the fact that taking both intergals PdV +VdP = PV, one obtains the change in internal energy of the system for process 1-2.

5. ### Q_Goest

2,985
Hi outrageous. I think you're confusing "state function" with "path function". See for example:
http://en.wikipedia.org/wiki/Functions_of_state
A fluid has some amount of enthalpy which is only dependent on it's physical state. Enthalpy is independent of how the fluid got to that state. For example, if we know the pressure and temperature of some gas, we know the physical state of that gas including it's density, internal energy, enthalpy and every other state function. You can get values such as density, internal energy and enthalpy from charts such as steam tables or from computer databases such as NIST. Note that these 'state functions' are also often called "thermophysical properties". So the work done by a fluid as it undergoes various changes in state does not depend on the initial and final states of the fluid - the work depends on the path taken by that fluid through the various physical states.

6. ### Outrageous

375
Thank Q_Goest and 256bits .
I think I understand that path and state function but what does this mean?

DrDu alludes to the fact that taking both intergals PdV +VdP = PV, one obtains the change in internal energy of the system for process 1-2.

7. ### Andrew Mason

6,880
I think 256bits meant to say: PdV + VdP = d(PV). d(PV) represents the change in internal energy only if dH = 0. In that case, then dH = dU + d(PV) = 0 so dU = -d(PV)

AM

8. ### DrDu

4,420
Did I really?
Well, obviously pdV=d(pV) when p is constant and thus int pdV is a change in state function, namely the change of pV. But what I wanted to stress is that pdV only equals the work in a quasi-static process. So W is not a state function even if p is hold constant.

9. ### Studiot

Heat transferred and work done across the system bounday (First Law; q and w) are never ever state functions. Period.

They cannot be as they can be determined by agencies outside of the system.

However they can sometimes be related to changes in system state functions because they produce changes in these.

Normally they are not proper differentials either. However under appropriate circumstances 'dq' and 'dw' can be regarded as such and integrated. this does not make them state functions under these circumstances

10. ### Outrageous

375
Thank you guys.

11. ### Zeppos10

84
in the preceding discussion no distinction is made between the pressure of the system and the pressure of the atmosphere. Further, no distincion is made between work done by/on the atmosphere and other work done by an external agent (called technical work). There is also work done on/by the system. Due to the failure to do so, the discussion is troublesome. See https://www.physicsforums.com/showthread.php?t=338573 for a different take.

12. ### Studiot

I followed your trail of links (quickly) as far as the pdf but could not see any reference to state functions.

13. ### Zeppos10

84
Discussions in this forum repeat themselves without much progress (my difficulty).
The question raised by "outrageous" was: So can I say work done on a system at constant pressure is a state function ? The answer to this question is yes. The state variable (the use of the word "function" has no function here) is called 'energy of displacement', the pressure involved is the external pressure. Enthalpy is the sum of the internal energy and the energy of displacement: no strings attached.

14. ### Studiot

Well since I've been here (a couple of years) I've seen students, come, study thermo and move on. Some have been outstanding, some less so. Perhaps there appears to be no progress because they move on to study other areas of physics and no longer post and they are replaced by a new intake of students that start again from the beginning?

This is more problematic.

Consider the following system, comprising a thin copper disk.

Applied to one face of the disk is an output shaft and a friction pad.
Applied to the other is a refrigerant wash and a control system set to maintain the temperature of the disk to any desired accuracy.

Thus the system so described satisfies both conditions asked by Outrageous - That of constant temperature and of constant pressure.

Within reasonable limits I can use the mechanism to input any quantity of work I desire and measure this directly as shaft work in BHP or other units.
Whatever value of work I input the state of the system does not alter so the input work is not a state function or state variable.

Last edited: Jan 3, 2013
15. ### DrDu

4,420
The energy of displacement is nothing else than pΔV which clearly is a state function at constant pressure. This has been amply discussed in this thread.
The problem is that this is usually not the only work done in an irreversible process, e.g. you can increase the internal energy of a fluid by stirring it. This change in internal energy can be compensated by removing heat. Hence any amount of work can be converted into heat even in a cyclic process whence work can not be a statefunction, even at constant pressure.
This is exactly how Benjamin Thompson (Count Rumford) arrived at his conclusion that heat cannot be a substance.

16. ### Studiot

Care needs to be taken in relating PdV or ∫PdV to first law work, w.

This is what the original Joule experiment (not the Joule-Thompson experiment which was later and different) was all about.

It is instructive to calculate the work done and ∫PdV for this situation.