00:01
Today, we're told that we have a sealed flask of 250 milliliters that contains 0 .55 grams of butane, and we're asked how much butane is present as a liquid.
00:10
We know that the heat of vaporization of butane is 22 .44 kilojoules per mole, and our normal boiling point is negative 0 .4 degrees celsius.
00:21
And so our end goal is to be able to use our yield gas equation pv equals nrt to solve for the number of moles of butane we have.
00:28
However, we don't yet know the pressure, so we cannot use the ideal gas equation yet.
00:34
Instead, we have to use the closest clapron equation to solve for our pressures because we know some temperatures.
00:41
We know that the flask contains this amount of butane at negative 22 degrees celsius, and that our normal boiling point is negative 0 .4 degrees celsius.
00:51
So let's convert these into kelvin.
00:53
This is going to be 272 .6 kelvin.
00:59
And this is going to be 251 kelvin.
01:07
In addition, this needs to be joules per mole, not kilojoules.
01:13
So we have 22 ,440.
01:18
And now we can use our closh's clapron equation, where we have the natural log is equal to p2 over p1, which is equal to negative delta h over our constant and r times one with our second temperature minus our first temperature.
01:41
And we can plug in the values that we notice all for our pressures...