00:01
For this problem on the topic of entropy, we had to suppose that 0 .55 moles of an ideal gas is isothermally and reversibly expanded in four of the situations given in the problem.
00:11
We want to find the change in entropy for each situation of this gas.
00:17
Now for an isothermal ideal gas process, we have the heat transfer queue equal to the work done w, and this is equal to nrt times the natural log of.
00:30
Of vf over vi, which means that the change in entropy, delta s, which is q over t, is w over t, which is n times r, times the natural log of vf over vi.
00:59
So if we apply this to the first scenario, we have the ratio of volumes, vf over vi, to be 0 .8 over 0 .2, which is equal to 4.
01:20
And therefore we have the change in entropy, delta s to be 0 .55 times 8 .31 joules per mole kelvin, which is the gas constant r times the natural log of 4, which gives us the change in entropy of 6 .34 joules per k.
01:43
So now if we do the same for part b, we have the ratio of volumes vf to vi to be 0 .8 over 0 .2...