# Reversible compression of a gas - faulty reasoning?

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kibestar
Hello,

I am trying to figure out where my reasoning falls apart in this thought experiment:

To determine if a process "A" is reversible (or at the very least internally reversible), I try to picture a reversible process "B" that involves only heat transfer and links the same two endpoints that bound process "A".

In this particular case, "A" is the reversible compression of a fixed mass of any one ideal gas confined in a frictionless piston-cylinder arrangement. Given that a reversible process by definition does not lead to entropy generation, the gas undergoes isentropic compression and receives an amount of energy equal to the PV work done by the piston upon the gas. State 1, then, can be characterized by entropy s1 = s, temperature T1 and volume V1. State 2 can be correspondingly be defined by s2 = s and temperature T2 != T1, and volume V2.

"B" is as follows: first, heat is removed reversibly from the system while the volume is kept constant until the pressure and temperature are only infinitesimally above absolute zero. At this point, the piston is allowed to move, changing the volume from V1 to V2. Once again the volume is kept fixed by some means and heat input is effected reversibly. Since the internal energy of an ideal gas is dependent only on its temperature, an amount of heat different from that which was removed at the beginning of the process must flow inward so as to attain temperature T2, which is different from T1. It follows that the amount of entropy transfer must also be different, which would entail different entropy values for the initial and final states, which is unverifiable in process "A". However, the volume V2 and temperature T2 are the same as those which lie at the end of process "A", which is the same as saying that both states are identical.

My question is: where have I made a mistake? Is it because the ideal gas hypothesis falls apart over this range of different temperatures/pressures? Or is it the assumption of constant specfic heat? If so, how does the underlying mathematics "know" this and reflect accordingly?

jartsa
So we have cooled the gas very close to absolute zero, and then the piston moves and compresses the gas, and increases the temperature of the gas by an infinitesimal amount - which might mean that the absolute temperature of the gas doubles - which is a significant change.

(It's significant because if temperature is very low, then a very small amount of heat will cause a significant increase of entropy)

(I don't know but I am guessing that the reason that we cool the gas so much is that it's supposed to make the effect of the motion of the piston insignificant)

Last edited:
kibestar
So we have cooled the gas very close to absolute zero, and then the piston moves and compresses the gas, and increases the temperature of the gas by an infinitesimal amount - which might mean that the absolute temperature of the gas doubles - which is a significant change.

(It's significant because if temperature is very low, then a very small amount of heat will cause a significant increase of entropy)

(I don't know but I am guessing that the reason that we cool the gas so much is that it's supposed to make the effect of the motion of the piston insignificant)

I believe it really comes to the non-zero temperature/pressure pair, as you pointed out. At first I had envisioned absolute zero temperature and pressure for this hypothetical process. However at that stage, the definition of entropy ceases to be mathematically sound. Lowering the temperature of the gas to absolute zero would call for an infinite amount of entropy being removed from the system. Likewise, raising its temperature from absolute zero would call for an infinite amount of entropy being transfered into the system. So I hypothesized infinitesimal temperature and pressure as a way of circumventing this "limitation". I guess I wasn't as rigorous as I thought I was being.