Question

How can we calculate `DeltaS` for an isothermal process?

Answer 1

Well, there are two ways I can think of. Both show that entropy increases as volume increases at constant temperature.

METHOD 1

You can always start from the first law of thermodynamics:

`DeltaE = q + w`

It is simplest to consider a reversible process, i.e. the 100% efficient process:

`DeltaE = q_(rev) + w_(rev)`,

where:

• `q_(rev)` is the reversible heat flow.
• `w_(rev) = -int PdV` is the reversible work done from the perspective of the system.
• `DeltaE` is the change in internal energy.

For an isothermal process, all ideal gases have `DeltaE = 0` and `DeltaH = 0`, since these are only functions of temperature for ideal gases. Thus,

`q_(rev) = -w_(rev) = int PdV`

Now, consider the definition of the entropy:

`dS = (deltaq_(rev))/T`,

where `delta` indicates a path function and `d` indicates a state function. Both are differentials and can be integrated.

Thus, by finding `q_(rev)`, one can find `DeltaS`. The change in entropy would have been:

`int_((1))^((2)) dS = int_((1))^((2)) 1/T deltaq_(rev)`

`= DeltaS = q_(rev)/T`

For ideal gases, we know the form of the pressure.

`P = (nRT)/V`

`=> q_(rev) = nRT int_(V_1)^(V_2) 1/V dV`

`= nRTln(V_2/V_1)`

So, we have:

`color(blue)(DeltaS("const. T") = nRln(V_2/V_1))`

METHOD 2

Write the total differential of the entropy as a function of temperature and volume:

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

The constant temperature process, or isothermal process, is the second term, so we seek only:

`DeltaS = cancel(int_(T_1)^(T_2) ((delS)/(delT))_VdT)^(0)` `+` `int_(V_1)^(V_2) ((delS)/(delV))_TdV`

`T` and `V` are natural variables in the Maxwell Relation for the Helmholtz free energy `A`:

`dA = -Sul(dT) - Pul(dV)`

Since all thermodynamic functions (`G, A, H, E, S, T, V, P, . . .`) are state functions (whose differentials are "exact"), the cross-derivatives here are equal:

`((delS)/(delV))_T = ((delP)/(delT))_V`

So, we have:

`DeltaS ("const. T") = int_(V_1)^(V_2) ((delP)/(delT))_VdV`

And for ideal gases, we again know the form of the pressure:

`P = (nRT)/V`

So its partial derivative with respect to temperature at constant volume is `nR//V`. This gives:

`color(blue)(DeltaS("const. T")) = nRint_(V_1)^(V_2) 1/VdV`

`= color(blue)(nRln(V_2/V_1))`

just as we got before.