# What is isothermal expansion of a real gas?

Sep 13, 2017

Well, isothermal expansion of any gas has $\Delta T = 0$, i.e. is at constant temperature.

In general, we may want to find $\Delta U$ and $\Delta H$, the changes in internal energy and enthalpy.

• For ideal gases, neither $\Delta U$ nor $\Delta H$ 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:

$\underline{\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} \left[T {\left(\frac{\partial P}{\partial T}\right)}_{V} - P\right] \mathrm{dV}}$

$\underline{\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} \left[- T {\left(\frac{\partial V}{\partial T}\right)}_{P} + V\right] \mathrm{dP}}$

As an example to show that these expressions hold for ideal gases, recall that $V = \frac{n R T}{P}$ and $P = \frac{n R T}{V}$. This means that the following partial derivatives are:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{n R}{V}$

${\left(\frac{\partial V}{\partial T}\right)}_{P} = \frac{n R}{P}$

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

$\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} T \cdot \frac{n R}{V} - \frac{n R T}{V} \mathrm{dV} = 0$

$\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} \left[- T \cdot \frac{n R}{P} + \frac{n R T}{P}\right] \mathrm{dP} = 0$

which shows that ideal gases have $\Delta U$ and $\Delta H$ as NOT functions of temperature.

DISCLAIMER: DERIVATION BELOW!

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

$\mathrm{dU} = T \mathrm{dS} - P \mathrm{dV}$

$\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}$

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

$\Delta U = {\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dU} = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV}$

$\Delta H = {\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dH} = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial H}{\partial P}\right)}_{T} \mathrm{dP}$

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

${\left(\frac{\partial U}{\partial V}\right)}_{T} = T {\left(\frac{\partial S}{\partial V}\right)}_{T} - P {\cancel{{\left(\frac{\partial V}{\partial V}\right)}_{T}}}^{1}$

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:

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$

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

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

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

${\left(\frac{\partial U}{\partial V}\right)}_{T} = T {\left(\frac{\partial P}{\partial T}\right)}_{V} - P$

And so, for ANY gas, we evaluate:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaU = int_(V_1)^(V_2) [T((delP)/(delT))_V - P] dV" }}{|}}$

Similarly, using the Maxwell Relation for the enthalpy:

${\left(\frac{\partial H}{\partial P}\right)}_{T} = T {\left(\frac{\partial S}{\partial P}\right)}_{T} + V {\cancel{{\left(\frac{\partial P}{\partial P}\right)}_{T}}}^{1}$

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

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$

and we get:

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$

which gives us:

${\left(\frac{\partial H}{\partial P}\right)}_{T} = - T {\left(\frac{\partial V}{\partial T}\right)}_{P} + V$

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

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaH = int_(P_1)^(P_2) [-T((delV)/(delT))_P + V]dP" }}{|}}$