Question

What is isothermal expansion of a real gas?

Answer 1

Well, isothermal expansion of any gas has `DeltaT = 0`, i.e. is at constant temperature.

In general, we may want to find `DeltaU` and `DeltaH`, the changes in internal energy and enthalpy.

• For ideal gases, neither `DeltaU` nor `DeltaH` are functions of temperature, and so those go to zero for ideal gases.
• For real gases, those are NOT zero.

I derive the following two relations further below:

`ul(DeltaU = int_(V_1)^(V_2) [T((delP)/(delT))_V - P] dV)`

`ul(DeltaH = int_(P_1)^(P_2) [-T((delV)/(delT))_P + V]dP)`

As an example to show that these expressions hold for ideal gases, recall that `V = (nRT)/P` and `P = (nRT)/V`. This means that the following partial derivatives are:

`((delP)/(delT))_V = (nR)/V`

`((delV)/(delT))_P = (nR)/P`

Then we get (realizing that the integral of zero is zero, and plugging in `P` and `V` as well):

`DeltaU = int_(V_1)^(V_2) Tcdot(nR)/V - (nRT)/V dV = 0`

`DeltaH = int_(P_1)^(P_2) [-Tcdot (nR)/P + (nRT)/P]dP = 0`

which shows that ideal gases have `DeltaU` and `DeltaH` as NOT functions of temperature.

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DISCLAIMER: DERIVATION BELOW!

There are Maxwell Relations for each of these functions in a thermodynamically-closed system (no mass or energy transfer):

`dU = TdS - PdV`

`dH = TdS + VdP`

Since we wish to be at constant temperature, it is most convenient to define:

`DeltaU = int_((1))^((2)) dU = int_(V_1)^(V_2) ((delU)/(delV))_TdV`

`DeltaH = int_((1))^((2)) dH = int_(P_1)^(P_2) ((delH)/(delP))_TdP`

From the Maxwell Relations, we can get these partial derivatives:

`((delU)/(delV))_T = T((delS)/(delV))_T - Pcancel(((delV)/(delV))_T)^(1)`

The entropy derivative is based on the natural variables `V` and `T`. These belong to the Helmholtz free energy and are shown in its Maxwell Relation:

`dA = -SdT - PdV`

Since it is a state function, the cross-derivatives are equal:

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

Thus, the internal energy derivative is able to be evaluated using gas laws:

`((delU)/(delV))_T = T((delP)/(delT))_V - P`

And so, for ANY gas, we evaluate:

`color(blue)(barul|stackrel(" ")(" "DeltaU = int_(V_1)^(V_2) [T((delP)/(delT))_V - P] dV" ")|)`

Similarly, using the Maxwell Relation for the enthalpy:

`((delH)/(delP))_T = T((delS)/(delP))_T + Vcancel(((delP)/(delP))_T)^(1)`

We similarly know the entropy derivative, by using the Maxwell Relation for the Gibbs' free energy, so we start with:

`dG = -SdT + VdP`

and we get:

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

which gives us:

`((delH)/(delP))_T = -T((delV)/(delT))_P + V`

and we get a form that can be evaluated using ANY gas law to model ANY gas:

`color(blue)(barul|stackrel(" ")(" "DeltaH = int_(P_1)^(P_2) [-T((delV)/(delT))_P + V]dP" ")|)`