Free expansion of an ideal gas.

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In a free expansion of an ideal gas, the work done is zero due to the absence of external pressure, leading to the question of whether heat (q) is also zero. The discussion emphasizes that while the internal energy (U) of an ideal gas depends solely on temperature (T), it is crucial to establish whether T remains constant during the process. Participants note that although the process is typically considered adiabatic, the lack of explicit mention leaves room for debate about heat exchange. Ultimately, the consensus suggests that without work being done or heat transfer occurring, q must indeed be zero, reinforcing the principles of the first law of thermodynamics. This clarification highlights the relationship between work, heat, and internal energy in thermodynamic processes.
scorpion990
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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.
 
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The problem is usually described as adiabatic free expansion of an ideal gas to address this issue.
 
It's the definition of an adiabatic course
 
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
 
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!
 
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