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

by Outrageous
Tags: constant, enthalpy, function, state, temperature, workat
 P: 365 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
 Sci Advisor P: 3,119 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.
P: 365
Do you mean that work done in a reversible process and at constant pressure , can be considered as a state function?

 Quote by DrDu Work becomes a state function for adiabatic processes where it defines the change of internal energy U.
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?

P: 1,115

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

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.
PF Patron
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Sci Advisor
P: 2,794
Hi outrageous. I think you're confusing "state function" with "path function". See for example:
http://en.wikipedia.org/wiki/Functions_of_state
 In thermodynamics, a state function, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state (independent of path). A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state. In contrast, mechanical work and heat are process quantities because their values depend on the specific transition (or path) between two equilibrium states. The opposite of a state function is a path function.
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.
 P: 365 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.
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P: 6,357
 Quote by Outrageous 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.
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
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P: 3,119
 Quote by 256bits 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.
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.
 P: 5,462 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
 P: 365 Thank you guys.
 P: 83 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 http://www.physicsforums.com/showthread.php?t=338573 for a different take.
P: 5,462
 Zeppos There is also work done on/by the system. Due to the failure to do so, the discussion is troublesome......for a different take.
What was the relevence and also your difficulty please?

I followed your trail of links (quickly) as far as the pdf but could not see any reference to state functions.
 P: 83 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.
P: 5,462
 Zeppos Discussions in this forum repeat themselves without much progress (my difficulty).
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?

 Zeppos 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.
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.
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P: 3,119
 Quote by Zeppos10 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.
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.
 P: 5,462 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.

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