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,
Everything is fine, until the last formula. That is not the correct expression for the work differential of an ideal gas
ResponderBorrarIf I just saw my notes and you are right excuse me I was confused with the instructions, in that case the interpretation of a would be the length l of the side of a cube and 3a would be the volume of the cube
ResponderBorrar** a^3 would be the volume of the cube
ResponderBorraryes precisely. So you got the correct expression for the work differential in terms of "a". But we want to show that it is equal to the expression for the work differential -- which everybody should remember -- in terms of the volume V = a^3.
ResponderBorrarvery well I will correct it and I will upload it
ResponderBorrar