Free expansion of an ideal gas.

In summary, the gas does no work and does not change temperature. This has implications on the internal energy of the gas.
  • #1
scorpion990
86
0
I know that the work done on the system in any free expansion is 0 since the external pressure is 0. However.. is q necessarily 0? Does the temperature necessarily stay constant for an ideal gas?

I've been trying to justify the reasons for which q is necessarily 0, but I can't find a reason... Does anybody know why this is so? I know that U varies only with T for an ideal gas, but that would require justifying that the change in temperature is also 0.. does anybody have a mathematical justification for why q is necessarily 0 for a free expansion of an ideal gas?

Thanks.
 
Physics news on Phys.org
  • #2
The problem is usually described as adiabatic free expansion of an ideal gas to address this issue.
 
  • #3
It's the definition of an adiabatic course
 
  • #4
scorpion990 said:
I know that the work done on the system in any free expansion is 0 since the external pressure is 0. However.. is q necessarily 0? Does the temperature necessarily stay constant for an ideal gas?

I've been trying to justify the reasons for which q is necessarily 0, but I can't find a reason... Does anybody know why this is so? I know that U varies only with T for an ideal gas, but that would require justifying that the change in temperature is also 0.. does anybody have a mathematical justification for why q is necessarily 0 for a free expansion of an ideal gas?

Thanks.
Apply the first law. Does the gas do work? Does heat flow into or out of the gas from/to the surroundings? What does that tell you about internal energy of the gas?

AM
 
  • #5
Andrew Mason said:
Apply the first law. Does the gas do work? Does heat flow into or out of the gas from/to the surroundings? What does that tell you about internal energy of the gas?

AM

The gas does no work, but heat could, in theory, flow in or out of the gas. I don't see why it can't... The problem doesn't specifically say that the process is adiabatic, but... I guess it is.. Obviously, that would solve all of the problems.

Thanks everybody!
 

What is free expansion of an ideal gas?

The free expansion of an ideal gas is a process in which a gas expands into a vacuum without any external work being done on the gas. This results in a decrease in the gas's pressure and temperature.

What causes free expansion of an ideal gas?

Free expansion occurs because of the second law of thermodynamics, which states that energy will spontaneously flow from areas of higher concentration to areas of lower concentration. Since the vacuum has no gas molecules, the gas will naturally expand into it.

What happens to the temperature and pressure during free expansion?

During free expansion, the temperature and pressure of the gas decrease. This is because the gas molecules are spreading out and taking up a larger volume, resulting in a decrease in both the number of collisions between molecules and the overall energy of the system.

Is free expansion reversible?

No, free expansion is an irreversible process. This means that the gas cannot spontaneously return to its original state without external work being done on it. This is due to the fact that the energy of the gas is dissipated into the larger volume of the vacuum.

What is the significance of free expansion in thermodynamics?

Free expansion is significant because it illustrates the second law of thermodynamics and the concept of entropy, which is a measure of the disorder of a system. It also demonstrates that the thermodynamic properties of a gas are dependent on its surroundings.

Similar threads

Replies
2
Views
565
  • Classical Physics
Replies
27
Views
892
  • Classical Physics
Replies
6
Views
789
  • Classical Physics
Replies
10
Views
1K
  • Classical Physics
Replies
14
Views
1K
Replies
14
Views
1K
Replies
18
Views
1K
Replies
5
Views
1K
Replies
0
Views
514
  • Classical Physics
Replies
5
Views
1K
Back
Top