Question

What is the entropy change for the isothermal expansion of `"0.75 g"` of `"He"` from `"5.0 L"` to `"12.5 L"`?

Answer 1

I got `"1.428 J/K"`.

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I assume you aren't given an equation and have to derive it.

If we treat entropy as a function of the temperature and volume (that is, `S = S(T,V)`), and consider a constant-temperature expansion, then the total derivative is:

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

Since we are looking at the change in entropy due to a change in volume, we only consider `((delS)/(delV))_T`. Take the time to digest this step. This is the key step to understanding where to begin.

Another key step is to relate this to a derivative we are more familiar with. Recall that the Helmholtz free energy `A` is a function of temperature and volume, and that its Maxwell relation is:

`dA = -SdT - PdV = -(SdT + PdV)`

Since `A` is a state function, its second derivatives are continuous, so that its cross derivatives are equal.

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

You should get to know this pretty well. The righthand side is a derivative we are familiar with. The ideal gas law can then be used.

`PV = nRT`

`=> P = (nRT)/V`

Now, the derivative is feasible in terms of an equation we know:

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

`= (nR)/V(dT)/(dT)`

`= (nR)/V`

This therefore gives:

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

Next, we can integrate this over the two volumes to get the change in entropy. By moving the `dV` in the `((delS)/(delV))_T` derivative onto the other side:

`int_(S_1)^(S_2) dS = DeltaS = int_(V_1)^(V_2) (nR)/VdV`

The integral of `1/V` is `ln|V|`, so we have:

`DeltaS = nRint_(V_1)^(V_2) 1/VdV = nR|[ln|V|]|_(V_1)^(V_2)`

`= nR(lnV_2 - lnV_1)`

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

So, given `"0.75 g"` of helium, we can get the mols of gas and thus find the change in entropy.

`color(blue)(DeltaS) = 0.75 cancel"g He" xx cancel("1 mol")/(4.0026 cancel"g He") xx "8.314472 J/"cancel"mol"cdot"K" xx ln("12.5 L"/"5.0 L")`

`=` `color(blue)("1.428 J/K")`