Question

What is the change in entropy for an adiabatic expansion against an external pressure for an ideal gas?

Answer 1

I assume you mean the entropy of the system. This won't be zero since the volume of the gas changes, assuming the expansion is reversible and that the gas is ideal.

---

If we are to examine `DeltaS_(sys)` for an adiabatic expansion, what we should recall is that:

• Reversible work is `w_(rev) = -int_(V_1)^(V_2) PdV`
• Reversible heat flow is `deltaq_(rev) = TdS`

In an expansion, the change in volume is positive, so that `w_(rev) < 0`. If it's adiabatic, the heat flow between the system and surroundings is zero, i.e. `deltaq_(rev) = 0`.

By the first law of thermodynamics:

`DeltaU = cancel(q_(rev))^(0) + w_(rev) = -P int dV`,

or in differential form:

`dU = cancel(deltaq_(rev))^(0) + deltaw_(rev) = -PdV`, `" "" "bb((1))`

where `U` is the internal energy.

So all the work goes into changing the internal energy of the system, while `DeltaS` due to reversible heat flow is zero.

However, the change in entropy can also be written as a function of temperature `T` and volume `V`:

`dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV` `" "" "bb((2))`

This treats the change in entropy as two steps:

1. Change in entropy due to a change in temperature at a constant volume.
2. Change in entropy due to a change in volume at a constant temperature.

Note that the second term involves a change in entropy due to a change in volume, which is related to the work. With some manipulation using Maxwell's Relations, we end up with:

`dS(T,V) = 1/T((delU)/(delT))_VdT + ((delP)/(delT))_VdV` `" "" "bb((3))`

where `P` is pressure, and `T` is temperature.

For an ideal gas, we can evaluate using the ideal gas law that

`((delP)/(delT))_V = (del)/(delT)[(nRT)/V]_V = (nR)/V`,

so:

`dS(T,V) = 1/T((delU)/(delT))_VdT + (nR)/VdV` `" "" "bb((4))`

Note that in an adiabatic process, we have that `dU = -PdV` from `(1)`, so dividing by `dT` at a constant volume:

`((delU)/(delT))_V = -Pcancel(((delV)/(delT))_V)^(0) = 0`

Treating the change in entropy in two steps, integrating `(4)` over a temperature range for the first step and volume range for the second step:

`DeltaS(T,V) = int_(T_1)^(T_2) 1/Tcancel(((delU)/(delT))_V)^(0)dT + int_(V_1)^(V_2) (nR)/VdV`

But the integral of zero is zero (no area under the curve!), so:

`color(blue)(DeltaS(T,V)) = 0 + color(blue)(nRln(V_2/V_1))`

Thus, `DeltaS ne 0` for an adiabatic expansion against an external pressure for an ideal gas.