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Is the formula only applicable during quasi-static processes? In other words, is it only true for a gaz at equilibrium?
For an ideal gas, I don't see how PV=nRT holds unless there is equilibrium. Pressure and temperature are defined only for equilibrium. The extreme case would be a free expansion of a ball of gas. While the gas is expanding, PV \ne nRT. Pressure is non-uniform (greatest at the centre and 0 at the edges). Since the kinetic energy of the gas molecules does not follow the Boltzmann distrbution, temperature of the gas as a whole is not really defineable.Clausius2 said:No, that equation holds even under translational, vibrational, rotational and chemical non equilibrium of the gas molecules. One can arrive to that equation from the Kinetic Theory and also from the Statistical Mechanics.
BUT, have into account that only ideal gases yield that equation. With that I mean that only small perturbations off the equilibrium are allowed.
quasar987 said:But when a piston abruptly compresses a gaz, pressure becomes non uniform in the gaz (it is higher near the piston surface since there is a net accumulation of particles there). So what is to be taken as P in the ideal gaz formula then? The higher P, the lower P? The average?
Andrew Mason said:For an ideal gas, I don't see how PV=nRT holds unless there is equilibrium. Pressure and temperature are defined only for equilibrium. The extreme case would be a free expansion of a ball of gas. While the gas is expanding, PV \ne nRT. Pressure is non-uniform (greatest at the centre and 0 at the edges). Since the kinetic energy of the gas molecules does not follow the Boltzmann distrbution, temperature of the gas as a whole is not really defineable.
AM
Clausius2 said:Maybe my assertion was too naive and vague. I should have complemented it with other corolary: the time of relaxation of the nonequilibrium process must be much shorter than the characteristic time of flow.