Free expansion of a Van der Waals gas, physical explanation

In summary, when a Van der Waals gas undergoes a free expansion, it cools down due to the attractions between its particles requiring heat energy from the surroundings to overcome during expansion. This heat is then released during compression. This is in contrast with an ideal gas, where the temperature remains unchanged during a free expansion as there are no attractions between particles. The change in temperature for a Van der Waals gas during free expansion is dependent on the attraction between particles and the change in volume, rather than the size of the particles. The change in temperature can be explained by the interplay between kinetic and potential energy of the gas, where the potential energy increases as the molecules move farther apart, resulting in a decrease in kinetic energy and temperature.
  • #1
fluidistic
Gold Member
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Hello!
Today I've learned that when a Van der Waals gas undergoes a free expansion, it cools down a bit. :bugeye:
This is in contrast with the ideal gas in which case since it does no work and since the process is adiabatic, the internal energy of the ideal gas remains unchanged by the expansion and since the temperature of the gas depends strictly on its internal energy, the temperature remains unchanged.
However for a Van der Waals gas, one has ##\Delta T=\frac{2an}{3R} \left ( \frac{1}{V_f} - \frac{1}{V_i} \right )##. Where a is, according to Wikipedia:
WikiTheGreat said:
a measure of the attraction between the particles
.
So the change of temperature doesn't seem to depend on the size of the gas' particles, only on the attraction between particles and the change in volume.

I don't really grasp it physically. What's going on for a Van der Waals gas during a free expansion? Why is the temperature going down?!
 
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  • #2
The conceptual explanation seems straight forward to me : If there are attractions between
the gas molecules then heat energy from the surroundings will be required to overcome them
during expansion. And of course that heat will be released during compression.
 
  • #3
morrobay said:
The conceptual explanation seems straight forward to me : If there are attractions between
the gas molecules then heat energy from the surroundings will be required to overcome them
during expansion. And of course that heat will be released during compression.

Ah bingo!
That makes sense, thank you.

Edit: Hmm I'm not really sure. If I'm not wrong for both the ideal and Van der Waals gas, the total energy remains unchanged, no heat is being released/absorbed from the surrounding during the free expansion and no work is done either.
I'm still at a loss.
 
  • #4
Imagine there is no change in internal energy (as the gas can expand into a vacuum in an adiabatic bath).

Consider the interplay between kinetic and potential energy of the gas.
 
  • #5
Jorriss said:
Imagine there is no change in internal energy (as the gas can expand into a vacuum in an adiabatic bath).
Yeah I noticed.
Joriss said:
Consider the interplay between kinetic and potential energy of the gas.

Hmm I'm not sure at all.
Apparently the kinetic energy decreases because the temperature decreases during the expansion? This would mean that the potential energy increases (is is the enthalpy?). But I don't get really understand why this happens.
Ah.. I think it's because the molecules are getting farther away from each other so the potential energy of "attraction" between the molecules increases. Since the total energy must remain constant, this means that the kinetic energy has to lower, and hence the temperature.
Does that sound correct?
 

1. What is the concept of free expansion in a Van der Waals gas?

Free expansion is a process in which a gas expands into an empty container without any external work being done on it. In a Van der Waals gas, this expansion occurs due to the random motion of gas molecules.

2. What is the physical explanation for the free expansion of a Van der Waals gas?

The physical explanation for free expansion in a Van der Waals gas is based on the kinetic theory of gases. This theory states that gas molecules are in constant random motion and collisions with each other and the container walls cause pressure. When the gas expands into an empty container, the molecules have more space to move around and collide with each other, resulting in a decrease in pressure.

3. How does the Van der Waals equation account for the free expansion of a gas?

The Van der Waals equation takes into account the attractive forces between gas molecules and the volume occupied by the molecules themselves. During free expansion, the gas molecules move further apart, reducing the attractive forces and the volume occupied by the molecules, resulting in a decrease in pressure.

4. Can free expansion occur in all types of gases?

Yes, free expansion can occur in all types of gases, including Van der Waals gases. However, the extent of free expansion may vary depending on the characteristics of the gas, such as the strength of intermolecular forces and the size of gas molecules.

5. Is free expansion an isothermal process in a Van der Waals gas?

No, free expansion in a Van der Waals gas is not an isothermal process. During free expansion, there is no heat exchange with the surroundings, so the temperature of the gas remains constant. However, there is a decrease in pressure and an increase in volume, which violates the definition of an isothermal process.

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