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## Main Question or Discussion Point

Assume we have an ideal gas of N particles inside a thermally isolated cylinder of volume V, and that the cylinder is equipped with a piston that can trap the air on one side. (Assume the piston occupies no volume in the cylinder.) Initially, the piston is fully withdrawn and the gas occupies a the entire volume V.

Step one:

The piston is slowly moved until the gas is compressed to half its original volume, V/2. Assume that this step is entirely quasistatic.

Step two:

A crack suddenly forms in the piston and the gas leaks back into the original volume V.

I am confused about what the entropy change is after step two. I can't seem to reconcile the thermodynamic answer with my statistical mechanics reasoning. Please help me out. Here are my thoughts:

Step one is explicitly stated to be quasistatic. Since the system is thermally isolated, I'm thinking that the process is adiabatic, since it is in equilibrium the entire time and no heat flows. For a reversible process, dS = dQ/T, and since dQ=0, we have that dS=0. So there is no entropy change during a quasistatic compression of a thermally isolated ideal gas. So the entropy change at the end of step one is zero.

Step two is Joule expansion. Joule expansion is not reversible and there is a nonzero entropy change. Since the energy E of an ideal gas is given by E = 3/2 NkT, and since no heat or work is performed on the system, the temperature must be the same. Since entropy is a function of state, one can say that the entropy change of the gas is equal to the entropy change along an isothermal expansion, which is given by [itex]\Delta S = \int_{V}^{2V} \frac{p}{T}dV = \int_{V}^{2V} \frac{Nk}{V}dV = Nk\ln(2)[/itex] where I have used the ideal gas equation of state PV=NkT.

So after step 2 is finished, there is a nonzero entropy increase equal to Nk ln(2). So my thermodynamic reasoning says there is an entropy change.

However, when I think about the statistical mechanics of this, I am baffled. The energy of the gas E=3/2 NkT never changes over the whole process, and thus after step two is over, the gas has returned to the same volume and energy it had before step 1. Thus the number of microstates of the gas after step 2 has to be exactly the same as the number of microstates before step 1. If we use the statistical definition of entropy, S = k lnΩ, then ΔS=0.

I guess that this second line of reasoning must be wrong, but I cannot figure out why. Would someone explain this to me?

Step one:

The piston is slowly moved until the gas is compressed to half its original volume, V/2. Assume that this step is entirely quasistatic.

Step two:

A crack suddenly forms in the piston and the gas leaks back into the original volume V.

I am confused about what the entropy change is after step two. I can't seem to reconcile the thermodynamic answer with my statistical mechanics reasoning. Please help me out. Here are my thoughts:

Step one is explicitly stated to be quasistatic. Since the system is thermally isolated, I'm thinking that the process is adiabatic, since it is in equilibrium the entire time and no heat flows. For a reversible process, dS = dQ/T, and since dQ=0, we have that dS=0. So there is no entropy change during a quasistatic compression of a thermally isolated ideal gas. So the entropy change at the end of step one is zero.

Step two is Joule expansion. Joule expansion is not reversible and there is a nonzero entropy change. Since the energy E of an ideal gas is given by E = 3/2 NkT, and since no heat or work is performed on the system, the temperature must be the same. Since entropy is a function of state, one can say that the entropy change of the gas is equal to the entropy change along an isothermal expansion, which is given by [itex]\Delta S = \int_{V}^{2V} \frac{p}{T}dV = \int_{V}^{2V} \frac{Nk}{V}dV = Nk\ln(2)[/itex] where I have used the ideal gas equation of state PV=NkT.

So after step 2 is finished, there is a nonzero entropy increase equal to Nk ln(2). So my thermodynamic reasoning says there is an entropy change.

However, when I think about the statistical mechanics of this, I am baffled. The energy of the gas E=3/2 NkT never changes over the whole process, and thus after step two is over, the gas has returned to the same volume and energy it had before step 1. Thus the number of microstates of the gas after step 2 has to be exactly the same as the number of microstates before step 1. If we use the statistical definition of entropy, S = k lnΩ, then ΔS=0.

I guess that this second line of reasoning must be wrong, but I cannot figure out why. Would someone explain this to me?

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