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:
.
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?!
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:
Quote:
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Quote by WikiTheGreat
a measure of the attraction between the particles
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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?!