00:01
We recognize that vaporization, when we are at the boiling point, corresponds to a normal boiling point in this case, corresponds to a system at equilibrium.
00:13
If a system is at equilibrium, then delta g is equal to zero.
00:21
If delta g is equal to zero, then according to the equation, delta g equals delta h minus t delta s, then we can solve for the molar standard change in entropy, 0, will be equal to the delta h of vaporization, 27 .49 kilojoules per mole, that's provided, minus the temperature in kelvin, its boiling point is 47 .6 degrees celsius, so we'll add 27 .13 .1 .7 .1 .7.
01:02
To get to that to get it into kelvin and we can now solve for delta s standard and delta s standard becomes 0 .0857 kilojoules per kelvin or multiplying that by a hundred by a thousand we get 85 .7 joules per kelvin and this of course course was per mole up here, so we still have our mole units.
01:41
Now knowing this, we can use what we calculated as a conversion factor when we have 28 .6 grams of the substance undergoing vaporization at its normal boiling point...
