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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.

flag	asked Nov 21 at 14:11 aceth 31●4

Greg P · Answer 1 · 2010-04-15 21:17:49Z UTC

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There is nothing for the freely expanding system to exchange heat with, right? So q = 0.

answered Apr 15 at 21:17

Greg P
116●3

Noldorin · Answer 2 · 2010-07-16 16:24:15Z UTC

You can't justify that Q = 0 unless you either assume either:

a) the expansion is isothermal and the working fluid is an ideal gas,

or

b) the expansion is adiabatic (which gives you Q = 0 immediately) - often considered by putting the system in isolated or contact with a heat reservoir.

I need help with free expansion of an ideal gas

2 Answers