00:01
For this problem on the topic of entropy, we have a mole of monotomic ideal gas being taken through the cycle given in the figure, where v1 is 4 v0, and we want to calculate the ratio w over p .0 v0 as the gas goes from state a to state c along path a, b .c.
00:16
We then want to find the ratio of the change in internal energy delta e over p .0 v0 as it goes from b to c, then through one full cycle.
00:27
We also want to find the change in entropy delta s in going from b to c and through one full cycle.
00:34
Now work is done only for the a -b portion of the process, and this portion is at constant pressure, so the work done by the gas, w is equal to the integral from v -0 to 4 -v -0 of p -0 -d -v, which is p -0 into 4 v -0 minus 1 -v -0.
01:02
Which is 3 p .0 v0, which means that the ratio that we require, w over p .0 v0 is equal to 3.
01:24
For part b, we use the first law, which gives us the change in internal energy to be the heat transfer queue minus the work done w.
01:36
And since the process is at constant volume, the work done by the gas is zero.
01:40
And so the internal energy is just q.
01:44
The energy q absorbed by the gas as heat is equal to n.
01:48
Cv delta t, where cv is the molar -specific heat at constant volume and delta t the change in temperature...