Correction to today's lecture

I think I made a mistake in today's lecture for the example we discussed involving sand being removed from the top of a piston. Here is a longer explanation that may be useful to you.

Recall that the two process we considered were:

(A) The sand is removed grain by grain. Each time a grain is removed, the piston inches up just a little bit higher as the gas expands a small amount.

(B) The sand is swept off instantly. The piston sharply rebounds and oscillates until it dampens out and comes to the same final state as for the process (A).

The work W is more negative for process A than for B. Recall that a negative work means the system does work on its environment. Thus, process A does more work on the environment than process B. The reason is because in A, the piston lifts much more sand against its weight--at each step, it's lifting an amount of sand that's slowly decreasing with time (as we remove grains). On the other hand, process B lifts only the residual sand that's left after we swept it off. Less sand means less gravitational force that is overcome, and hence less work.

Both the final states are the same for the two processes. Why? It is because the pressure of the gas is the same at the end of A and B (they have the same amount of sand on top of the piston) and because the temperature is the same (we're at constant temperature in this example.) That's 2 intensive variables for a single-component, single-phase system, which means the equilibrium state of the gas is completely specified by the Gibbs phase rule. That means that the change in internal energy for both processes--which is a state function--is the same, since the beginning and ending equilibrium states are the same.

The first law says that Delta U = Q + W. Since Delta U is the same for A and B and since W is less (more negative) for A, the heat transferred Q must be greater for A than B in order to achieve the same value of Delta U. (This was stated as the opposite in class).

You can rationalize this physically: when a gas expands, it tends to cool. Therefore, to bring it back to the same temperature, heat must be transferred to it. That heat comes from the environment. In the case of B, all of that lost work that we didn't use to raise the sand was used (by way of friction) to heat the gas back up to the same temperature. So case B needs less heat from the environment to counter the effect of cooling upon expansion.

MSS