00:01
Hi, so here we're mixing two different gases that are correspondingly.
00:05
First of all, we have nitrogen gas that has a pressure equal to 2 .1 bar, the volume is going to be equal to one liter, and we're going to mix it up with argon gas that is at a pressure of 3 .4 bar and for a volume of four liters.
00:32
So in the first case, the temperatures are the same.
00:35
Let's say that temperature is going to be a constant in this scenario.
00:43
So what we know is that we're going to mix it up in a final volume equal to two liters.
00:52
So what do i need to do in both of these cases? remember that the pressure at the final is going to follow an ideal gas law, right? so this is going to be nrt divided by my final volume.
01:14
So in the first case, temperature is the same.
01:21
In the second case, temperature of nitrogen is going to be equal to 304 kelvin.
01:32
Temperature of argon is going to be equal to 402 kelvin.
01:38
And temperature of the mix is going to be equal to 377 kelvin.
01:53
Okay, so i forgot to hit play.
01:58
So essentially, i don't know where this is stopped, but the result over here of the mix, it is going to be 7 .85 bar.
02:06
We calculated the amount of moles that we're going to get.
02:11
We cancel everything that was possible.
02:14
And we found that the pressure of the mix is going to be pressure of nitrogen, volume of nitrogen plus pressure of argon, volume of argon divided by the volume of the mix.
02:24
Doing that calculation gives 7 .85 bar.
02:26
So now, in the second case, we're going to employ the same formula for the amount of mole.
02:33
But now, in this amount of mole, we can do the direct calculation.
02:39
So we are going to plug in the value for the temperature of nitrogen that is 304 kelvin, giving us the result.
02:49
With this, we are going to get 2 .1 times 1 divided by 0 .082 divided by 304, giving us an amount of mole of 0 .08424...