Thermodynamics of a fast process

Question

The given question states that. N atoms of a perfect gas are contained in a cylinder with insulating walls, closed at one end by a piston. The initial volume is V and the initial temperature T. (a)Find the change in temperature, pressure and entropy that would occur if the volume were suddenly increased to V2 by withdrawing the piston.

Now the given solution says that The gas does no work when the piston is withdrawn rapidly. Also, the walls are thermally insulating, so that the internal energy of the gas does not change, i.e., dU = 0. Since the internal energy of an ideal gas is only dependent upon temperature T, the change in temperature is 0, i.e., Tz = TI. As for the pressure, p2/p1 = Vl/V2.

I was really puzzled by the given solutions, first of all it says that the gas does no work, I am not sure how this is possible if the volume is changing, second it says that as the walls are thermally insulating the internal energy does not change, I am pretty sure that thermally insulated walls imply that dQ=0, not dU=0.

The given problem is from MIT, I am really not sure what is going on

Answer 1

first of all it says that the gas does no work, I am not sure how this is possible if the volume is changing

The only way to remove energy from an insulated, expanding perfect gas is for the molecules to bounce off a retreating wall, losing momentum with every collision. If the wall movement is slow, this happens many, many times. If the wall movement is sufficiently fast, it might not happen even once.

You have to interpret "if the volume were suddenly increased" to mean a sufficiently fast withdrawal that no molecule has a chance to collide with the retreating wall. This is equivalent to free expansion (sometimes also visualized as the rupture of a barrier separating a gas from vacuum). Since $W=0$ and $Q=0$ from the insulation, $\Delta U=0$.

Answer 2

The issue with this poorly worded problem are the words "suddenly withdrawn." What the solution that they present implies is that the (tensile) force per unit area that you are applying to suddenly withdraw the piston is exactly equal in magnitude and opposite in direction to the outside atmospheric pressure throughout the expansion process. This means that, throughout the process, the net external force on the massless frictionless piston is zero, and thus, no net work is done on (or by) the gas. So W = 0, and since Q is also equal to zero, we have that $$\Delta U=Q-W=0$$For an ideal gas, this means that the gas temperature is constant.

Answer 3

Consider the mass of the gas as a closed thermodynamic system, that is one whose mass does not change, but whose boundary moves as the volume of the gas increases (same mass, different volume). For the expansion process, the first law applied to the system is $Q_{in} - W_{out} = \Delta U$, where $Q_{in}$ is the heat into the system (the fixed mass of gas), $W_{out}$ the work done by the system, and $\Delta U$ is the change in the internal energy $U$ of the system.

This is an irreversible, non-isentropic process since rapid expansion is not a quasi-equilibrium process; therefore, you cannot evaluate $W_{out}$ as $\int_{V_{initial}}^{V_{final}} P(V)dV$ where $P$ is the gas pressure and $V$ the gas volume. (You can evaluate $W_{out}$ as $\int_{V_{initial}}^{V_{final}} P(V)dV$ for a quasi-equilibrium process where the gas progresses through a set of thermodynamic states.) See my answer to Work done by a gas on this exchange. However, you can always consider $W_{out}$ as the work done by the gas as it expands against external forces, here the pressure in the volume created by withdrawal of the piston. The gas is not expanding against the piston, it is expanding into the vacuum created when the piston is suddenly withdrawn to a new position (assumed to be an "instantaneous" repositioning of the piston), then held in that position. Since the gas expands into a vacuum, the gas is pushing against zero pressure, and $W_{out}$ is zero. $Q_{in}$ also given as zero. Based on the first law $\Delta U$ is zero and the internal energy of the gas does not change. For an ideal gas, the internal energy (and also the enthalpy) are functions of temperature only, so for an ideal gas undergoing this process there is no change in temperature. Real gases are not ideal and experiments have shown that the temperature decreases slightly (fractions of a degree F) in this process.

This is the famous Joule experiment of 1843, a free expansion process, discussed in all basic thermodynamics textbooks. For example, see one of the books on thermodynamics by Sonntag and Van Wylen.

Answer 4

Assume the piston is massless and that the pressure of the ideal gas is initially in equilibrium with the external atmospheric pressure.

I could be wrong, but I think what they are saying is that the work done by the external agent pulling on the piston is not done on or by the enclosed gas, but is done by the external agent against the atmosphere as its pressure will be greater than the gas pressure during the withdrawal and thus oppose the withdrawal of the piston.

Insofar as the ideal gas goes, when the piston is pulled back by the external force, the gas molecules simply move further apart to fill the increased volume. Since, for an ideal gas, it is assumed there are no intermolecular forces holding the molecules together and that all the internal energy is kinetic, no work would be required by the gas to move the molecules apart.

Hope this helps.