# What is isothermal expansion of a real gas?

##### 1 Answer

Well, **isothermal expansion** of any gas has

In general, we may want to find **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

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

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

Then we get (realizing that the integral of zero is zero, and plugging in

#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

**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

#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" ")|)#