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I am reading Arieh Ben-Naim's book "A Farewell to Entropy", trying to understand "indistinguishable particles".
What if you have a gas composed of billiard balls and their collisions are perfectly elastic with each other and with the walls of their continer. Each ball has a unique serial number marked on it, but these marks do not change the collision physics. You can read the numbers, but the process of reading them does not appreciably affect the trajectory or collision processes.
Now at equilibrium the balls will have a Maxwellian velocity distribution. You can make all the usual thermodynamic measurements on this gas - you measure pressure as the average force on a large area, the temperature will be 2E/3k where E is the average energy per ball and k is Boltzmann's constant. Will the thermodynamics of this gas be qualitatively different than that of a container of say, hydrogen gas? In particular, if you measure the entropy (or entropy change) of this gas by thermodynamic methods, will it be extensive? I mean, the particles are distinguishable, and its always said that this means the entropy is not extensive.
I think that the answer is yes, the entropy will be extensive. Since "thermodynamic measurements" does not include shining a beam of light on individual balls, and since you cannot read the serial number by thermodynamic methods, the balls are "thermodynamically indistinguishable" and the entropy will be extensive. Does this sound right?
What if you have a gas composed of billiard balls and their collisions are perfectly elastic with each other and with the walls of their continer. Each ball has a unique serial number marked on it, but these marks do not change the collision physics. You can read the numbers, but the process of reading them does not appreciably affect the trajectory or collision processes.
Now at equilibrium the balls will have a Maxwellian velocity distribution. You can make all the usual thermodynamic measurements on this gas - you measure pressure as the average force on a large area, the temperature will be 2E/3k where E is the average energy per ball and k is Boltzmann's constant. Will the thermodynamics of this gas be qualitatively different than that of a container of say, hydrogen gas? In particular, if you measure the entropy (or entropy change) of this gas by thermodynamic methods, will it be extensive? I mean, the particles are distinguishable, and its always said that this means the entropy is not extensive.
I think that the answer is yes, the entropy will be extensive. Since "thermodynamic measurements" does not include shining a beam of light on individual balls, and since you cannot read the serial number by thermodynamic methods, the balls are "thermodynamically indistinguishable" and the entropy will be extensive. Does this sound right?