Lecture 19: Cooling real gas

Note

Stay cool and be happy. – Anonymous

Warning

This lecture corresponds to Chapter 27 of the textbook.

Summary

Attention

In the previous lecture (Lecture 18: Real gas), we introduced more realistic ways to describe a gas. The natural question is thus: what difference does this more realistic description have on the thermal properties? In this lecture, we focus on ways to cool a realistic gas.

1. Joule Expansion: Our first stop along the way is our old friend the Joule expansion. We know now that if an ideal gas is used, its temperature cannot change during the process (since it is adiabatic and does not involve work). This is a direct consequence of the fact that the internal energy of an ideal gas only depends on temperature (that makes sense since a change in volume/pressure could bring the molecules closer or farther away from one another but since they do not interact, the energy does not change. Conversely, a change in temperature leads to a change in speed and, in turns, to the kinetic energy).

What happens with a real gas? To quantify this, we introduce the Joule Coefficient:

$\mu_{ J }=\left(\frac{\partial T}{\partial V}\right)_{U}.$

This quantity is computed at constant internal energy since, even for a real gas, the energy is constant for the Joule expansion.

Now, we remember all the work we did when we studied thermodynamic potentials and this allows us, after using the reciprocal theorem and a Maxwell relation, to write:

$\mu_{ J }=-\frac{1}{C_{V}}\left[T\left(\frac{\partial p}{\partial T}\right)_{V}-p\right].$

In addition, one can write the change in temperature during the process as:

$\Delta T=\int_{V_{1}}^{V_{2}} \mu_{ J } d V=-\int_{V_{1}}^{V_{2}} \frac{1}{C_{V}}\left[T\left(\frac{\partial p}{\partial T}\right)_{V}-p\right] d V.$

Using these equations, we confirm that for an ideal gas, the Joule coefficient is zero, as expected. However, for a real gas, the temperature is found to go down! How do we understand this?

We remember that the Joule expansion takes place at constant internal energy. During the expansion, molecules get farther and farther away, thus leading to an increase in potential energy. To compensate this increase and keep the total energy constant, the kinetic energy of the molecules has to go up, which results in an increase in temperature.

2. Joule-Kelvin expansion: in contrast to the Joule expansion, the JK expansion is performed at constant enthalpy. In addition, what is varied is the pressure of the gas, not its volume.

As shown in the figure on the left, if the internal energy on the left and on the right are $U_1$ and $U_2$ , the change in energy due to the work done to move $V_1$ at pressure $p_1$ and to move $V_2$ at pressure $p_2$ is

$\Delta U= U_{2}-U_{1}=p_{1} V_{1}-p_{2} V_{2},$

in other words:

$U_{1}+p_{1} V_{1}=U_{2}+p_{2} V_{2}.$

We remember that the enthalpy is simply:

$H=U+pV,$

and therefore:

$H_1=H_2.$

The Joule-Kelvin expansion is thus an isenthalpic process!

The change in temperature during the process is described by the so-called Joule-Kelvin coefficient:

$\mu_{ JK }=\left(\frac{\partial T}{\partial p}\right)_{H},$

and we find:

$\mu_{ JK }=\frac{1}{C_{p}}\left[T\left(\frac{\partial V}{\partial T}\right)_{p}-V\right].$

The temperature changes as:

$\Delta T=\int_{p_{1}}^{p_{2}} \frac{1}{C_{p}}\left[T\left(\frac{\partial V}{\partial T}\right)_{p}-V\right] d p.$

It is more difficult to know if the process leads to heating or cooling. How the system behaves (as we move from high to low pressure) depends on what initial conditions we have. This process is explained in detail in the screencast.

Note

Entropy of a real (vdW) gas

The entropy of a real gas can be calculated from the equation of state. We find:

$S=C_{V} \ln T+R \ln (V-b)+\text { constant }.$

The remarkable result is that the entropy only depends on $b$ (and not $a$ ). This makes sense when we remember that the entropy calculates the number of possible microstates (that is: the number of ways to arrange the molecules). This number is dictated by the size of each molecule, which is exactly what the parameter $b$ defines.

Key Definitions

Note

Joule-Kelvin expansion:: The expansion of a gas through a small opening or a porous plug with the pressure on both side being maintained at two different pressures $p_1$ and $p_2<p_1$ . The expansion is also known as the Joule-Thomson expansion – since Thomson was Kelvin’s name before he was ennobled in 1892.
Joule coefficient:: Coefficient providing a quantitative knowledge on how the temperature changes during a Joule expansion. Its mathematical expression is:

$\mu_{ J }=\left(\frac{\partial T}{\partial V}\right)_{U}.$
Joule-Kelvin coefficient:: Coefficient providing a quantitative knowledge on how the temperature changes during a Joule-Kelvin expansion. Its mathematical expression is:

$\mu_{ JK }=\frac{1}{C_{p}}\left[T\left(\frac{\partial V}{\partial T}\right)_{p}-V\right].$
Inversion curve:: In a Joule-Kelvin expansion, the inversion curve is the curve when the isenthalps experience a change of slope in the $p(T)$ space. This means that the system goes from cooling to warming (or vice-versa) when crossing that curve.
Isenthalpic process:: Process taking place to constant enthalpy. Example: the Joule-Kelvin process is isenthalpic.

A full list of terms, including the ones provided here, can be found in the Index.

Learning Material

Copy of Slides

The slides for Lecture 19 are available in pdf format here: pdf

Screencast

Test your knowledge

Imagine that a real gas is used in a Joule expansion experiment (a Joule expansion occurs when a gas expands against vacuum in a thermally isolated system). Among all the statements below, which one is the most accurate?
1. The internal energy decreases, thus cooling.
2. The internal energy increases, leading to a decrease in kinetic energy, thus cooling.
3. The internal energy is constant and the interaction between the gas molecules increases, thus heating.
4. The internal energy is constant and the attractive interaction between the gas molecules increases, thus cooling.
5. The internal energy is constant and the attractive interaction between the gas molecules decreases, leading to an increase in energy and a corresponding decrease in kinetic energy, thus cooling.
6. The internal energy is constant and the interaction between the gas molecules decreases, leading to a decrease in potential energy and a corresponding increase decrease in kinetic energy, thus heating.
An ideal gas is used for a Joule expansion experiment, therefore:
1. The internal energy is constant, and the gas cools down
2. The internal energy is constant, and the gas heats up
3. The internal energy goes down, and the gas cools down
4. The internal energy goes down, and the gas heats up
5. The internal energy is constant, and the gas remains at the same temperature
Under otherwise similar conditions ( $p$ , $V$ , and $T$ ), is the entropic content of an ideal gas larger, smaller, or the same as that of a real gas?
1. Same
2. Larger
3. Smaller
4. It depends
During an isothermal expansion of a gas:
1. The internal energy of the gas always increases, even for an ideal gas.
2. The internal energy of the gas always decreases, even for an ideal gas.
3. The internal energy of the gas always remains constant, even for an ideal gas.
4. The internal energy always increases for a real gas.
5. The internal energy always decreases for a real gas.
6. The internal energy always remains the same for a real gas.
Consider the Joule-Kelvin expansion of a real gas.
1. It is always possible to find a pressure so that the gas cools down, regardless of the starting temperature.
2. It is not always possible to find a pressure so that the gas cools down, regardless of the starting temperature.
3. There isn’t enough information to answer this question conclusively.
The Joule-Kelvin expansion is a process that takes place…
1. at constant temperature
2. at constant Gibbs energy
3. at constant enthalpy
4. at constant entropy
5. at constant Helmholtz free energy
What does the real gas model account for that the ideal gas model neglects?
1. Superposition of particle states, and therefore the interactions between particles
2. The non-zero particle size, and interactions between particles
3. The ideal gas law does not neglect anything, hence ‘ideal’
4. The size of the container and the number of particles

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the following problems from the textbook: