00:05
So with the following equation, we can take a look at what we've actually got to do.
00:11
So we've got dp over dt equals delta h vaporization over temperature that is multiplied by vg minus vi.
00:29
So we assume that vi is equal to zero.
00:34
So then the equation becomes the following.
00:47
So we can rearrange the equation using vg equals nrt.
00:54
Over p and then we can substitute that in for dp over p equals delta h vaporization d t over nr t squared so now we can substitute in the value of delta h vaporization into our equation that is 15 .971 add 14 .55 subtract 0 .160 t squared so that would look as follows we have d p over p equals 15 .971 add 14 .55 t subtract 0 .160 t squared or multiplied by the differential of our temperature there divided by r t so then this is equal to dp over p equals 15 .971 over rt squared, add 1 .5 .5t over rt squared again, subtract 0 .162t squared over rt squared, all in square brackets, multiplied by dt.
02:39
So we simplify the above equations and then we can integrate from p1 to p2 as well as t1 to t2.
02:49
So like i said, we integrate in between p1, p2, and between t1, t2.
03:08
So then we have ln, p2 over p1 equals 15 .971 over r, multiplied by 1 over t1, subtract 1 over t2, add 1 .55 over r, ln, which is the natural logarithm, t2 over t2, p1, subtract 0 .1 .160 over r, which is 8 .314, multiplied by t2, takeaway t1.
03:50
So moving on to the second part that we have here...