- #1

Philip Koeck

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I've written a short text adapted from a previous post by Count Iblis. I'll append it below. It shows that irreversible PV-work is always smaller than reversible, which fits very nicely with W = P

Here comes the text:

The pV-work done by the system on the surroundings in an irreversible expansion is always smaller than the pV-work done in a reversible process between the same initial and final state.

In an irreversible compression the work done by the surroundings on the system is larger than in a reversible compression between the same states.

To see why this is the case, consider the first law of thermodynamics.

dU = dQ – dW (1)

In case of a reversible change of the system, we know that

dQ = T dS (2)

and

dW = P dV (3)

Inserting (2) and (3) in (1) gives:

dU = T dS - P dV (4)

which is known as the fundamental thermodynamic relation. Now, even though this was derived assuming the change in the system was reversible, it is also valid for irreversible changes. The reason is that the inner energy U is a thermodynamic state variable which can be uniquely specified by the entropy and volume. Then, if you change the volume and entropy, the internal energy change is some fixed quantity, independent of how that change is realized (reversible or not).

The fundamental thermodynamic relation (4) is just the first law of thermodynamics (1) written only with state variables and functions.

For irreversible changes we know that (2) is not valid.

In general we have

T dS ≥ dQ (5)

Inserting (4) on the left and (1) on the right of (5) and cancelling dU gives

P dV ≥ dW (6)

So, the work done in irreversible processes will, in general, be less than the integral of P dV.

Of course, if both dW and P dV are negative, as for a compression, this means that the absolute value of dW will be larger than the absolute value of P dV for compressions.

??Without proof we state that the PV-work in an irreversible expansion or compression at constant external pressure is given by

W = P

Notice that for P

_{ext}ΔV. I'd be interested if there's a way to show that W is exactly equal to P_{ext}ΔV for all expansions and compressions where the content of the cylinder qualifies as an ideal gas (pressure not too high and not too low). Another interesting thing would be experiments where this formula for work has been checked, maybe via the temperature change during a sudden expansion or compression.Here comes the text:

**PV-work for irreversible processes**The pV-work done by the system on the surroundings in an irreversible expansion is always smaller than the pV-work done in a reversible process between the same initial and final state.

In an irreversible compression the work done by the surroundings on the system is larger than in a reversible compression between the same states.

To see why this is the case, consider the first law of thermodynamics.

dU = dQ – dW (1)

In case of a reversible change of the system, we know that

dQ = T dS (2)

and

dW = P dV (3)

Inserting (2) and (3) in (1) gives:

dU = T dS - P dV (4)

which is known as the fundamental thermodynamic relation. Now, even though this was derived assuming the change in the system was reversible, it is also valid for irreversible changes. The reason is that the inner energy U is a thermodynamic state variable which can be uniquely specified by the entropy and volume. Then, if you change the volume and entropy, the internal energy change is some fixed quantity, independent of how that change is realized (reversible or not).

The fundamental thermodynamic relation (4) is just the first law of thermodynamics (1) written only with state variables and functions.

For irreversible changes we know that (2) is not valid.

In general we have

T dS ≥ dQ (5)

Inserting (4) on the left and (1) on the right of (5) and cancelling dU gives

P dV ≥ dW (6)

So, the work done in irreversible processes will, in general, be less than the integral of P dV.

Of course, if both dW and P dV are negative, as for a compression, this means that the absolute value of dW will be larger than the absolute value of P dV for compressions.

??Without proof we state that the PV-work in an irreversible expansion or compression at constant external pressure is given by

W = P

_{ext}ΔV, which fulfills (6)??Notice that for P

_{ext}= 0 the PV-work becomes 0, just as it should for a free expansion.
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