Paul and Andrea were discussing a thermally isolated free expansion of an ideal
ID: 1589341 • Letter: P
Question
Paul and Andrea were discussing a thermally isolated free expansion of an ideal gas. Recall that a free expansion is one where a gas expands into a vacuum. This is an adiabatic process. No transfer of heat occurs between the system and its environment and no work is done on or by the system. Andrea went first.
Paul replied.
“That's absurd. Imagine a free expansion into a very large volume. Let’s assume that we start with the gas at 1 atm and room temperature. Now let the gas expand into a space so large that after the expansion, in any cubic meter, the probability of finding a single molecule of the gas is vanishingly small. Do you mean to tell me, that if I were to stick a thermometer into this dilute gas after the expansion, it would read anything but zero? There arn’t any molecules around to strike my thermometer and transfer thermal energy to it. So, during the process the system has gone from room temperature to absolute zero. Temperature can’t be a constant for this process.”
Resolve this debate.
Explanation / Answer
We know that q = 0 for the expansion because the system is insulated. We also know that w = 0, because the gas is expanding into a vacuum (w = -PextV = 0 x V = 0). This leads us to the conclusion that the change in internal energy E = q + w = 0 + 0 = 0.
The internal energy (E) of the system is the sum of the of its potential energy (PE) and kinetic energy (KE). If E = PE + KE, then E = PE + KE. For E to be zero, the sum PE + KE must be zero too. In other words, the changes in KE and PE must be equal and opposite to each other.
But, for an ideal gas (and this is the key point), the potential energy is always zero. Potential energy is the energy of atoms interacting with other atoms, e.g., to make a chemical bond, or to attract or repel each other through non-covalent interactions. But in an ideal gas the molecules are non-interacting. This means PE is always zero for an ideal gas. Note that this is not true for a real gas. Real gases have attractions and repulsions between molecules that make PE 0.
E = 0 = PE + KE = 0 + KE, which means KE = 0. But we know from the kinetic-molecular theory that KE = 3nRT/2 for a monatomic ideal gas, so T = 2KE/3nR = 0. For a polyatomic ideal gas, KE is still proportional to T, and T = 0. This means that the temperature does not change in the expansion, and, if that is the case, we can now remove the insulation wtihout changing the result, since q = 0 for an isothermal system.
The solution then, is: q, w, E, PE, KE, and T are all zero.