Assignment "Tarea 06"
Oh, and I would like to remind you to the homework "Tarea 06" which is due for this Friday. If you have any questions, doubts, etc. please post to this blog.
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 have a question, in the problem T06.3, is only one expansion curve and only one adiabatic curve? , or we need complete the cycle for come back to A state?.
ResponderBorrarI have a similar question for the problem T06.3, the curve means that we have first a rect line, then a line tangent to this fist line, and finally a curve that begins in our final point "B" and finishes at our starting point "A", so we have something similar to a triangle for this graph?
ResponderBorrarSorry for the late response. In problem 6.3 we first have an expansion which is represented by a straight line and then an adiabatic compression which takes the system back to its original state.
ResponderBorrarIt should be clear that you can obtain a cycle with these two processes. What may be causing the confusion is the fact that the system absorbs heat only along some part of the straight-line-process, and it passes heat to the environment only at some other part of it.