00:02
Here we're going to take a stab at problem number 12 from chapter 12, where we've been once again handed a flask of gas.
00:09
They keep doing that to us, but it's okay we know what to do.
00:12
We've got a volume of 135 milliliters at 22 .5 degrees celsius.
00:18
Oh, is it not in kelvin? and that's exerting a pressure of 165 millimeters of mercury on the walls of our flask.
00:27
And we are tasked with transferring this to a new flask with different conditions.
00:32
Bam, bigger flask? 252 milliliters.
00:36
It is cooled on down to zero degrees celsius.
00:40
That makes me sad.
00:43
And we've been tasked to find what is the new pressure in this situation? to find this answer, we're going to once again, just like problem eight, make use of the combined gas law.
00:57
Pressure times volume divided by temperature.
01:00
Of our first situation equal to the pressure, volume, and temperature of our second situation.
01:10
In kelvin, we'll come to y in just a moment.
01:16
Let's not dilly -dally any time and just plug our measurements straight in there.
01:21
If you'd like to see the derivation of our combined gas law, you can look at the video for problem number eight.
01:28
Here we have a pressure of 165.
01:31
Actually let's do this down here 165 millimeters of mercury times our volume of 135 milliliters all this divided by our temperature which is 22 .5 converted to gelvin by adding 273 we set this equal to our mystery pressure what could it be times our new expanded volume 252 milliliters divided by our new pressure, 0 .0, excuse, 3 celsius, which we need to convert to kelvin.
02:13
It's especially obvious here.
02:15
If we don't, we're dividing by zero.
02:18
That's a spike in your graph.
02:19
How many nothings can you fit into something? you don't want to have to answer that question for your gas problem.
02:25
So we convert to kelvin when we are making use of the gas laws.
02:30
From here to isolate our pressure and find our answer, we just need to revisit our algebra...