Question

How do you calculate `DeltaS_(sys)` for the isothermal expansion of 1.5 mol of an ideal gas from 20.0 L to 22.5L?

Answer 1

I got `"1.47 J/K"`. This should make sense to be positive since a larger volume allows more particle movement, which increases the extent of energy dispersal in the system.

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We begin by realizing that an isothermal expansion is one where `DeltaT = 0`. The change in entropy due to a change in volume at a constant temperature is denoted as `((delS)/(delV))_T`.

To get from `V_1` to `V_2`, we integrate the following:

`DeltaS_"sys" = int_(V_1)^(V_2) ((delS)/(delV))_TdV`

We denote molar entropy as `barS = S/n` and molar volume as `barV = V/n`, so that we then have our starting equation:

`color(green)(DeltabarS_"sys" = int_(barV_1)^(barV_2) ((delbarS)/(delbarV))_TdbarV)`

From the Maxwell Relation for `barA(T,barV)` (the molar Helmholtz Free Energy as a function of temperature and molar volume)

`dbarA = -barSdT - PdbarV`,

we use the cross-derivative relationship to get

`((delbarS)/(delbarV))_T = color(green)(((delP)/(delT))_barV)`,

meaning that the change in entropy due to the change in volume at constant temperature is equal to the change in pressure due to the change in temperature at a constant volume.

Next, we would have to figure out what this new derivative gives for the ideal gas law. Recall that the ideal gas law is:

`PV = nRT`

or

`PbarV = RT`

`=> P = (RT)/(barV)`

The partial derivative of pressure with respect to temperature at constant molar volume is then:

`=> ((delP)/(delT))_barV = (del)/(delT)[(RT)/(barV)]_barV`

`= R/(barV)(del)/(delT)[T]_barV = R/(barV)((delT)/(delT))_barV`

`= R/(barV)cancel((dT)/(dT))`

`= R/(barV)`

So, what's left is to integrate the result by substituting `R/(barV)` back in for `((delbarS)/(delbarV))_T`:

`color(blue)(DeltabarS_"sys") = int_(barV_1)^(barV_2) R/(barV)dbarV`

`= Rint_(barV_1)^(barV_2) 1/(barV)dbarV = R|[ln(barV)]|_(barV_1)^(barV_2)`

`= R[ln(barV_2) - ln(barV_1)] = Rln(barV_2/(barV_1)) = Rln((V_2"/"cancel(n))/(V_1"/"cancel(n)))`

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

Notice how the molar volumes have canceled out to give the regular volume.

Therefore, the change in entropy for the system at constant temperature is:

`color(blue)(DeltaS_"sys") = nDeltabarS_"sys" = nRln(V_2/V_1)`

`= ("1.5 mols")("8.314472 J/mol"cdot"K")ln(("22.5 L")/("20.0 L"))`

`=` `color(blue)("1.47 J/K")`