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Free expansion of a Van der Waals gas, physical explanation

by fluidistic
Tags: expansion, explanation, free, physical, waals
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fluidistic
#1
May7-13, 08:37 PM
PF Gold
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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.
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:
Quote Quote by WikiTheGreat
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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morrobay
#2
May7-13, 09:04 PM
P: 376
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.
fluidistic
#3
May7-13, 09:26 PM
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Quote Quote by morrobay View Post
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.

Jorriss
#4
May7-13, 09:41 PM
P: 1,042
Free expansion of a Van der Waals gas, physical explanation

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.
fluidistic
#5
May7-13, 10:01 PM
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Quote Quote by Jorriss View Post
Imagine there is no change in internal energy (as the gas can expand into a vacuum in an adiabatic bath).
Yeah I noticed.
Quote Quote by Joriss
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?


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