00:01
Before we start the calculations of this problem, we should take note of a couple of the values that we're going to need.
00:07
So we're given that the pressure in atm is 520 tor.
00:13
We're also told that the enthalpy of vaporization is 40 .7 kilojoules per mole.
00:20
So what we need to know is that the boiling point of water is 100 degrees celsius, and the standard state condition, or in other words, the normal pressure, and tor is 760.
00:36
So we are going to be using this equation.
00:39
The natural log of our first pressure divided by our second pressure is equivalent to the enthalpy of vaporization divided by r, which is a constant, multiplied by one over our second temperature minus one over our first temperature.
00:57
And what we're trying to figure out is this value, 1 over t2.
01:01
So now we simply plug in all of the values we got before.
01:06
So on the left side of our equation, we'll get the natural log of 760 divided by 520.
01:14
And we'll set this equal to 40 ,700 joules.
01:18
Notice how we converted kilojoules to joules, divided by r, 8 .314 joules per mole kelvin, times 1 over t, that's what we need to know, minus 1 over 373.
01:32
Kelvin...