1.3: Understanding Real Gases Using Measurable Quantities

Introduction

Combining thermodynamics with measurable quantities allows us to predict the the thermodynamic behavior of real substances not just ideal ones. This significantly extends the usefulness of thermodynamics. In this section we will consider some examples of how real gases behave differently than ideal gases.

Adiabatic volume changes

For an ideal gas we have shown that \(\pi_T = 0\). This allows us to state that \(dU = \pi_T dV + C_V dT  = C_V dT\). \(dU\) is also \(dU = dq +dw\), but dq = 0 under adiabatic conditions so combining these we have rigorously shown that for an ideal gas:

\[C_V dT = dw.\label{adiab_w_ideal}\]

whereas for a real gas where \(\pi_T \ne 0\) the relationship is:

\[\pi_T dV + C_V dT  = dw.\label{adiab_w_real}\]

Case 1: Against zero pressure (irreversible)

If the gas expands against zero pressure it does no work and dw = 0.

Ideal gas case

Equation \(\ref{adiab_w_ideal}\) becomes:

\[C_V dT = 0.\nonumber\]

As \(C_V \gt 0\) this implies that dT = 0, so the gas does not change temperature.

Real gas case

Equation Equation \(\ref{adiab_w_real}\) becomes:

\[\pi_T dV + C_V dT  = 0. \nonumber\]

This rearranges to:

\[\pi_T dV = -C_V dT.  \nonumber\]

Collecting the measured quantities on one side and the differentials on the other yields:

\[\left(\frac{dT}{dV}\right)_S=-\frac{\pi_T}{C_V}.  \nonumber\]

In this case the subscript S is used to indicate adiabatic (isentropic) conditions. \(C_V \gt 0\) and in the density range where attractions dominate \(\pi_T \gt 0\). Thus the temperature will drop when a real gas expands against a vacuum. If \(\pi_T \lt 0\), the situation at very high densities where repulsions dominate, the temperature could increase upon expansion.

This result can also be written in terms of the easily measurable \(\alpha\) and \(\kappa_T\) using our previous result, \( \pi_T = T \frac{\alpha}{\kappa_T}-P \), resulting in:

For a real gas expanding against a vacuum: \[\left(\frac{dT}{dV}\right)_S=-\frac{1}{C_V}\left(T \frac{\alpha}{\kappa_T}-P\right).  \nonumber\]

Case 2: Against a non-zero pressure (irreversible)

In this case \(dw = -P_{ext}dV\), where \(P_{ext}\) is a constant external pressure that either allows expansion or causes compression.

Ideal gas case

Equation \(\ref{adiab_w_ideal}\) becomes:

\[C_V dT = -P_{ext}dV\nonumber\]

which rearranges to:

\[\left(\frac{dT}{dV}\right)_S = -\frac{P_{ext}}{C_V }\label{ideal_dT_adiab_Pext}.\]

So an expansion of an ideal gas against a constant pressure will result in a temperature drop. The rate of drop is determined by the ratio of the external pressure and the constant volume heat capacity of the gas.

Real gas case

Equation \(\ref{adiab_w_real}\) becomes:

\[\pi_T dV + C_V dT  = -P_{ext}dV. \nonumber\]

Collecting like differentials on the same sides produces:

\[(\pi_T +  P_{ext})dV = -C_V dT. \nonumber\]

Collecting the differentials on the left and other factors on the right under adiabatic conditions:

\[\left(\frac{\partial T}{\partial V}\right)_S = -\frac{\pi_T +  P_{ext}}{C_V} . \label{real_dT_adiab_Pext}\]

In the regime where \(\pi_T \gt 0\) the numerator in equation \(\ref{real_dT_adiab_Pext}\) is more positive than in equation \(\ref{ideal_dT_adiab_Pext}\) by \(\pi_T\). Thus, a real gas would be expected to cool more on expanding against a constant pressure than an ideal gas.  If \(\pi_T \lt 0\) then the real gas would cool less than an ideal gas. As in the previous instance we can substitute in for \(\pi_T\) in terms of \(\alpha\) and \(\kappa_T\).

For a real gas expanding adiabatically against a constant pressure: \[\left(\frac{\partial T}{\partial V}\right)_S = -\frac{\left(T \frac{\alpha}{\kappa_T}-P\right) +  P_{ext}}{C_V} .\nonumber\]

Case 3: Reversible volume change

 In the previous two examples we considered how temperature changed during an irreversible change in volume. During a reversible adiabatic change the pressure and volume must change in sync. A good way to understand this is to evaluate the partial derivative

\[ \left(\dfrac{\partial V}{\partial P }\right)_S \nonumber \]

This can be expanded by using the cyclic relation followed by an application of the reciprocal relation:

\[ \left(\dfrac{\partial V}{\partial P }\right)_S = - \dfrac{ \left(\dfrac{\partial V}{\partial S}\right)_P }{ \left(\dfrac{\partial P}{\partial S }\right)_V }. \nonumber \]

Expanding the two derivatives on the right-hand-side makes the equation easier to interpret:

\[ \left(\dfrac{\partial V}{\partial P }\right)_S = - \dfrac{ \left(\dfrac{\partial V}{\partial T}\right)_P \left(\dfrac{\partial T}{\partial S}\right)_P}{ \left(\dfrac{\partial P}{\partial T}\right)_V \left(\dfrac{\partial T}{\partial S}\right)_V}. \label{eq20} \]

Remembering that \(dS = \frac{dq_{rev}}{T}\), which means at constant P \(dS = \frac{C_P dT}{T} \implies \left(\frac{\partial S}{\partial T}\right)_P = \frac{C_P}{T}\) and likewise at constant V \(\left(\frac{\partial S}{\partial T}\right)_V = \frac{C_V}{T}\), equation \ref{eq20} can be simplified to:

\[ \left(\dfrac{\partial V}{\partial P }\right)_S= - \dfrac{C_V}{C_P} \left(\dfrac{\partial V}{\partial T}\right)_P \left(\dfrac{\partial T}{\partial P}\right)_V. \nonumber \]

\[ = \dfrac{C_V}{C_p} \left(\dfrac{\partial V}{\partial P}\right)_T \label{eq22} \]

Or defining \(\gamma = C_p/C_V\), Equation \ref{eq22} can be easily rearranged to

\[ \gamma \left(\dfrac{\partial V}{\partial P}\right)_S =\left(\dfrac{\partial V}{\partial P}\right)_T = -V\kappa_T\label{dVdP_adiab} \]

Ideal gas case

For an ideal gas \(\kappa_T = P^{-1}\) (at ambient conditions \(\kappa_T \approx 10^{-5} Pa^{-1}\)). So, substituting into equation \(\ref{dVdP_adiab}\) gives us:

\[ \gamma \left(\frac{\partial V}{\partial P}\right)_S =   -\frac{V}{P} \nonumber \]

which is in a form we can rearrange to integrate. Separation of variables yields:

\[ \gamma \frac{dV}{V} = -\frac{dP}{P}. \nonumber \]

Integration (assuming that \(\gamma\) is independent of volume) yields

\[ \gamma \int_{V_1}^{V_2} \frac{dV}{V} = -\int_{P_1}^{P_2} \frac{dP}{P}. \nonumber \]

Integrating both sides yields

\[ \gamma \ln \left( \frac{V_2}{V_1} \right) =-\ln \left( \frac{P_2}{P_1} \right) \quad \text{or} \quad \gamma \ln \left( \frac{V_2}{V_1} \right) =\ln \left( \frac{P_1}{P_2}\right),\nonumber \]

which is easily manipulated to show that

\[P_1V_1^{\gamma} = P_2V_2^{\gamma} \nonumber \]

or

\[PV^{\gamma} = \text{constant} \nonumber \]

the behavior of an ideal gas along an adiabat.

Real gas case

In this case we do not have a closed form expression for \(\kappa_T\), although we could use an empirical gas equation such as the virial expansion, van der Waals or others.

Consider again equation \(\ref{dVdP_adiab}\) but rearranged to put all the constants on one side:

\[ \left(\frac{\partial V}{\partial P}\right)_S = -\frac{V\kappa_T}{\gamma}\nonumber \]

In the case of real gases intermolecular attractions dominate except at high density (low T and/or high P). Imagine increasing the pressure of a "low" density real gas. Increasing the pressure increases the density, which lowers the average potential energy (closer together, so more attraction). In an adiabatic system the total energy (potential + kinetic) must remain fixed. If the potential energy is reduced then the kinetic energy must increase, which increases the pressure. Thus, a given change in pressure requires less of a change in volume. The conclusion is that \(\kappa_T(real) \lt \kappa_T(ideal)\) and that we see a smaller volume change for a given pressure change with real gases.

Closed form expressions for how P and V vary for significant changes in real gases are difficult to come up with because of the temperature dependence of \(\kappa_T\), which is often only available with good accuracy via direct measurements. For example consider the van der Waals empirical real gas equation:

\[\left(P + \frac{a}{\bar{V}^2}\right)\left(\bar{V}-b\right) = RT, \label{vdW}\]

where a and b are the van der Waals coefficients dependent on the identity of the gas and \(\bar{V}\) is the molar volume of the gas at the particular P and T combination. This is a cubic equation in \(\bar{V}\) so it is difficult to solve for \(\bar{V}\) and then take the derivative with respect to P to find \(\kappa_T\). However, we can use the reciprocal properties of derivatives to avoid having to do as much work, yet get a functional form of \(\kappa_T = \frac{-1}{V}\left(\frac{\partial V}{\partial P}\right)_T\). The trick is to realize we can solve easily for P and then just invert the derivative \(\left(\frac{\partial P}{\partial V}\right)_T\) to get a valid expression for what we want as illustrated below.

\[P=\frac{R T}{\bar{V} - b} - \frac{a}{\bar{V}^{2}}\nonumber\]

Taking the derivative yields:

\[\left(\frac{\partial P}{\partial \bar{V}}\right)_T=- \frac{R T}{\left(\bar{V} - b\right)^{2}} + \frac{2 a}{\bar{V}^{3}}.\nonumber\]

Inverting yields:

\[\left(\frac{\partial \bar{V}}{\partial P}\right)_T=\frac{\bar{V}^{3} \left(\bar{V} - b\right)^{2}}{- R T \bar{V}^{3} + 2 a \left(\bar{V} - b\right)^{2}}.\nonumber\]

Notice the temperature dependence. In an adiabatic process the temperature also changes with the volume, so a functional form of T(P, V) must be found in order to integrate over a significant volume change for a real gas.

Isothermal changes

Isothermal changes are what \(\kappa_T = - \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T\) and \(\pi_T=\left( \frac{\partial U}{\partial V} \right)_T\) quantify. As shown previously from the general result

\[ \pi_T = \left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{\alpha}{\kappa_T}-P, \nonumber \]

we know that for an ideal gas \(\pi_T = 0\), and, since \(\alpha\) and \(\kappa_T\) are defined to be positive, that \(\pi_T \gt 0\) for real gases unless the pressure is very high.

This gives us a handle on how U changes under isothermal  (dT = 0) conditions:

\[dU = dq + dw = \pi_T dV + C_V dT = \pi_T dV. \nonumber\]

Ideal gas case (\(\pi_T = 0\)): dU = 0, which implies \(\Delta U\) = 0 for all isothermal changes of an ideal gas.

Real gas case: \(dU \propto dV\). Thus, in the "low" pressure regime where \(\pi_T \gt 0\), U increases with increasing volume and decreases with decreasing volume. In the "high" pressure regime where \(\pi_T \lt 0\), the opposite is true.