Muddy Points for Lecture T3
1) There were many questions about the idea of process time versus the equilibration time. The point I was trying to make was as follows. We would like to deal with systems as if they were in equilibrium, in other words as if the system were in a static state with nothing changing. If they were in equilibrium, the pressure, temperature, etc. would be uniform throughout the system. To have the actual situation be a good approximation to this (to have the state changes be quasi-equilibrium or quasi-static) the time scale over which any change to the system occurs should be "much" longer than the time it takes the situation to come to equilibrium when disturbed. For example, if the time to come to equilibrium for the balloon is roughly 10R/a, this means that appreciable changes in the external conditions need to occur over a time scale "much" longer than 10R/a for the response of the system to be considered quasi-static. The word "much" is in quotes because the definition depends on what use we are making of the answer; in other words do we need something correct to 1% or something correct to 0.001%.
We cannot calculate the time to come to equilibrium from thermodynamics. The estimates come from information we have from other fields such as gas dynamics or heat transfer.
2) There were also many questions about the last example, where the external pressure and the system pressure were not equal because of friction. The solid line, which I pointed to as the system pressure, Ps, was supposed to be a line of constant temperature. For an ideal gas, PV = NRT. Therefore with constant temperature and the same amount of gas, PV = constant is the equation of the line. This describes what happens to the system, the stuff inside the chamber. The other lines on the diagram were the work received by us (the class!) when the piston is moved out, or the work that we must do on the system to move the piston in. Considering the direction of the frictional forces, and doing a force balance, you can see that we get less work than that calculated by using dW = PsdV when the system is moving out. The external pressure will also be lower than the system pressure when the piston is moving out (the gas in the system is expanding). To compress the gas the opposite is true. The work is greater than that given by dW = PsdV and the external pressure is higher than the system pressure when the gas is being compressed.
The message here is that to connect the work we (i.e. some entity outside the system) receive to changes in system properties, that is to be able to say dW = PsdV, we need:
(1) the state change to occur over a time which is long enough for the system to always be very nearly in equilibrium (i.e. for the system to act as if it were in equilibrium to a very good approximation) and
(2) the frictional forces to be negligible.