# First law of thermodynamics and quasi-static processes

## Main Question or Discussion Point

why is

du = δq - δw valid for all processes while

du = δq - p dv valid only for quasi-static processes ?

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Andrew Mason
Homework Helper
why is

du = δq - δw valid for all processes while

du = δq - p dv valid only for quasi-static processes ?
P refers to the internal pressure of the gas. But P does not necessarily determine the work done.

In an expansion, the work done by the gas is determined by the external pressure on the gas, not the internal gas pressure. If they are both equal (the external pressure is infinitessimally less than the internal pressure) the process will be very slow - quasi-static.

It is a little more difficult to see this point with a compression. One would think that the work done on the gas is determined by the internal gas pressure so dw = Pint dV even if it is not quasi-static. But if the external pressure is much greater than the internal gas pressure the gas molecules are given more kinetic energy than if the external and internal pressures were close to being equal. So dw = PdV for a compression only if it is quasi-static too.

AM

thank you very much Andrew !

i got the point.
i have few other doubts. i'll post them soon.
:)

m little bit confused. i hope you understand my question.

du = δq - δw

is valid also for irreversible process.

is it true that δq and δw values have to be given else we cannot determine du unless the process is reversible so that we can use integral p dv and integral T ds ?

i mean, if we are given the values of properties at point 1 and 2 in a reversible process, we can evaluate δq and δw using integral p dv and integral T ds.

so how do we determine δq and δw when the process is irreversible?

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Andrew Mason
Homework Helper
m little bit confused. i hope you understand my question.

du = δq - δw

is valid also for irreversible process.

is it true that δq and δw values have to be given else we cannot determine du unless the process is reversible so that we can use integral p dv and integral T ds ?
U is a state function so you don't need to know dW and dQ to determine dU. You just need to know the initial and final states.

If the path is reversible you just need to know the initial and final states (P,V,T) to determine the work done and heat flow in moving between those two states.

But if the path is not reversible, you need to know the initial and final states AND the work done in order to determine the heat flow (or the heat flow in order to determine the work done).

i mean, if we are given the values of properties at point 1 and 2 in a reversible process, we can evaluate δq and δw using integral p dv and integral T ds.

so how do we determine δq and δw when the process is irreversible?
The simple answer is that you can't. There are an infinite number of different irreversible processes that can cause a system to go from one state to another, each using different amounts of work and resulting in different heat flows to the surroundings. ΔU = ΔQ - W is always true, but you need to know either ΔQ to find W or W to find ΔQ if the process is not reversible.

[Size=+1]Correction: see post #7 below[/Size]

AM

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got it.

thanks once again Andrew!!
:)

Andrew Mason
Homework Helper
I have looked at this again and I have to correct my previous post #5. I said:
Andrew Mason said:
If the path is reversible you just need to know the initial and final states (P,V,T) to determine the work done and heat flow in moving between those two states.
This is only correct if you are using the most direct reversible path (eg. an isothermal path, adiabatic path).

There are many different reversible paths between any two states, each of which will involve different heat flows and work. You cannot determine the work done and heat flow involved in a reversible path or for an irreversible path if all you have are the beginning and end states. The actual path is needed.

For reversible paths W = ∫PdV (W = work done BY the gas) and ΔQ - W = ΔQ - ∫PdV = ΔU.

For irreversible paths W ≠ ∫PdV so ΔQ - ∫PdV ≠ ΔU. However: ΔQ - W = ΔU still applies.

But if the path is not reversible, you need to know the initial and final states AND the work done in order to determine the heat flow (or the heat flow in order to determine the work done).
You can determine the work done and heat flow for a reversible path if you are also given the path or the relationship between P and V during the process. For example, you can determine the change in internal energy and work done by an ideal gas undergoing an adiabatic reversible expansion between two states.

The real difference between reversible and irreversible paths is that the Work done by the system is not ∫PdV if the process is irreversible. So for an irreversible path, you need more than the relationship between P and V. You need to know the actual work done and/or heat flow.

AM

work done by/on the system on/by the surroundings is equal to
integral of (external pressure i.e, pressure of surroundings ) * (change in volume).

but if the process is quasi-static we can use internal pressure of gas (pressure of the system) instead of exteral pressure coz pint = pext ± dp, where dp is very very small.

does that mean du = δq - p dv (where p is internal pressure ) valid for reversible process while
du = δq - p dv (where p is external pressure) valid for both reversible and irreversible processes ?

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Andrew Mason
Homework Helper
work done by/on the system on/by the surroundings is equal to
integral of (external pressure i.e, pressure of surroundings ) * (change in volume).

but if the process is quasi-static we can use internal pressure of gas (pressure of the system) instead of exteral pressure coz pint = pext ± dp, where dp is very very small.

does that mean du = δq - p dv (where p is internal pressure ) valid for reversible process while
du = δq - p dv (where p is external pressure) valid for both reversible and irreversible processes ?
It is not quite as simple as that. In irreversible processes the system and immediate surroundings are not in thermodynamic equilibrium so the states are not precisely defined (eg. P and T). Thermodynamics deals with equilibrium states and transitions from one equilibrium state to another. PV=nRT only applies to an ideal gas in thermodynamic equilibrium, for example.

To analyse the work done in an irreversible process you would have to take into account the dynamic energies of the system and the surroundings and that can be difficult.

AM

Ken G
Gold Member
One way to think about this is that PdV is only dW for quasi-static situations (i.e., gases evolving from one equilibrium state to another). If both the gas inside and outside are at the same P and evolve quasi-statically, then the distinction between the work done by the interior gas and the work done by the exterior gas gets blurred because it is not an important distinction to make when they are the same (except for sign). But the distinction is always there-- we should never say that the work is done on the exterior gas rather than by the interior gas-- instead, we should simply note that both the interior and exterior gases are doing work, and we will need to analyze that work separately if we do not have a reversible quasi-static process.

I'm really not sure where the idea comes from that the work done by an irreversible process is the PdV on the external gas, that just isn't consistent use of language (I'm not saying anyone on here made up that language, you see it in a lot of places). Even for an irreversible process, there is a work done by the internal gas, and a work done by the external gas, and they can be different. Whether or not either one will be PdV just depends on whether or not that particular gas is maintaining an isotropic pressure, or if there is significant dynamical energy appearing that cannot be treated as quasistatic.

Apparently the case people often talk about in this context is so-called "free expansion", where you have an interior gas at much higher pressure than the exterior gas, and there it is clear that the interior gas is going to carry a lot of dynamical bulk-flow energy that cannot be treated as PdV work on its surroundings because P will not completely characterize the momentum flux density. But the surrounding gas should also generate significant bulk flows in this case, so PdV isn't going to be right for the work done by the external gas either unless the external gas has a much higher sound speed than the internal gas (so a shock won't be driven into the external gas). I think a lot of what is said about free expansion just doesn't make a whole lot of sense.

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