Question

How do you calculate the change in entropy for an isothermal process?

Answer 1

It depends on which way we change the entropy (pressure or volume?).

Pick one, and you would use one of these:

`DeltaS = nRln(V_2/V_1)`

`DeltaS = -nRln(P_2/P_1)`

Actually, if you notice that at constant `T` and `n` for a monatomic ideal gas, `P_1V_1 = P_2V_2`:

`ln(V_2/V_1) = ln((P_1cancel(V_1)"/"P_2)/(cancel(V_1)))`

`= ln[(P_2/P_1)^(-1)] = -ln(P_2/P_1)`,

which means our two expressions are equivalent for any monatomic ideal gas.

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DISCLAIMER: This contains Calculus. I assume that if you read further, you understand partial derivatives, cross-derivatives, and the integral of `1/x`.

Entropy is typically considered a function of temperature and either volume or pressure. When we hold temperature constant (an isothermal process), and change one of the other parameters:

(1) `DeltaS = int_(P_1)^(P_2) ((delS)/(delP))_TdP`

(2) `DeltaS = int_(V_1)^(V_2) ((delS)/(delV))_TdV`

If you notice, we don't have an expression for the way entropy changes due to pressure or volume in relation to gases. It would be nice to have a way to get these relationships from known formulas, like the ideal gas law.

It turns out that these can be derived from the Maxwell Relations that contain `T,P` or `T,V` as the variables that could potentially change:

(3) `dG = -SdT + VdP`,

where `G` is the Gibbs' free energy, and

(4) `dA = -SdT - PdV`,

where `A` is the Helmholtz free energy.

ENTROPY AS A FUNCTION OF TEMPERATURE AND PRESSURE

The first context in which we calculate `DeltaS` for an isothermal process involves looking at the cross-derivatives in (3); here is how you should examine `S,P,T`:

`" "" "" "uarr "------------------------------------" darr`
`dG = stackrel("Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) + V" "stackrel("Denominator")overbrace(dP)`
`" "" "" "" "" "" "" "uarr "---------------------" darr`

Do something similar for `V, T, P` to get:

`((delS)/(delP))_T = -((delV)/(delT))_P`

Then, for a monatomic ideal gas, we can use the ideal gas law (`PV = nRT`) to give an explicit expression for the partial derivative of `V` with respect to `T` at a constant `P`:

`V = (nRT)/P`

`=> color(green)(-((delV)/(delT))_P) = -(del)/(delT)[(nRT)/P]_P`

`= -1/P(d)/(dT)[nRT] = color(green)(-(nR)/P)`

That means our first way to get the change in entropy is (plugging back into (1)):

`color(blue)(DeltaS) = int_(P_1)^(P_2) ((delS)/(delP))_TdP`

`= -int_(P_1)^(P_2) ((delV)/(delT))_PdP`

`= -int_(P_1)^(P_2) (nR)/PdP = -nRint_(P_1)^(P_2) 1/PdP`

`= color(blue)(-nRln(P_2/P_1))`

ENTROPY AS A FUNCTION OF TEMPERATURE AND VOLUME

The other way to express `DeltaS` involves equations (2) and (4):

`" "" "" "uarr "------------------------------------" darr`
`dA = stackrel("Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) - P" "stackrel("Denominator")overbrace(dV)`
`" "" "" "" "" "" "" "uarr "---------------------" darr`

Do something similar for `P, T, V` to get:

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

On the ideal gas law, this means:

`P = (nRT)/V`

`=> color(green)(((delP)/(delT))_V) = (del)/(delT)[(nRT)/V]_V`

`= 1/V(d)/(dT)[nRT] = color(green)((nR)/V)`

Substitute this result back into (2) to get:

`color(blue)(DeltaS) = int_(V_1)^(V_2) ((delS)/(delV))_TdV`

`= int_(V_1)^(V_2) ((delP)/(delT))_VdV`

`= int_(V_1)^(V_2) (nR)/VdV = nRint_(V_1)^(V_2) 1/VdV`

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

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Overall, if you know the initial and final pressure, or the initial and final volume, you can figure out the change in entropy of a monatomic ideal gas at a constant temperature.