Solution to problem T05.3(c) @Sebastian
Hi Sebastian can you please post your solution to problem T05.3(c).
Sorry in a previous post I asked Yannick by mistake!
Sorry in a previous post I asked Yannick by mistake!
Work for an ideal gas in a 3D container (corrected)
Using the differential al work calculated on exercise 6.2: 1. How can we use the result to obtain DW for an ideal gas in a 3D container? What is the interpretation of the parameter "a" in that case. Solution. The differential of work calculated on exercise 6.2 is Here, d represents the dimension of the box. So The parameter a is the lenght, according to equation (67). Because we are working on 3 dimensions, a³=V . This might come from the definition of entropy*. 2. What do we get for DW for d=3 and n --> infinity? And how is this result related to the expected form DW = N k_B T dV/V? Solution. On the last expression we get DW for three dimensional case. Now, applying the limit on infinity, we expect that the differential of work would become on well potential one. So,
I didn't get the answer, I just made the exercises T05.1 and T05.2. I tried to do it with the past answer of the exercise T04.2, using an "if" but I didn't finished it.
ResponderBorrarOk, no problem.
ResponderBorrar