Here in this example question, it says a sample of gas initially has a volume of 900 milliliters at 520 Kelvin and 1.85 atmospheres. What is the pressure of the gas if the volume decreases to 330 milliliters while the temperature increases to 770 Kelvin?

All right. So in this question, they're giving me a volume. We're going to say it's V₁ because later on they give me a second new volume V₂. They give me a temperature T₁ and then later give me a second temperature T₂. They give me this pressure in atmosphere. So this is P₁ because later they ask me what is the new pressure. So they're asking us to find P₂.

Now we have our ideal gas law PV = nRT. If you have watched my videos on ideal law derivations, you know that we can manipulate the ideal gas law in order to get our combined gas law. Now, if you haven't watched those videos, I highly suggest you go back and take a look. So look under ideal gas Derivation videos.

Now here we're talking about two pressures. We're talking about two volumes. We're talking about two temperatures. Moles aren't being discussed because they're being held constant. R is a constant. We divide out the T so that everything is on the left side, and we see that since we're dealing with two different values for these variables, it becomes

P 1 V 1 T 1 = P 2 V 2 T 2

So here we just showed how we derived the combined gas law.

All right, so now we're going to plug in the values that we have. Our pressure is 1.85 atmospheres initially. Our volume, when we change the milliliters to liters, it gives us 0.900 liters. Remember, temperature must always be in kelvins when doing calculations. It's already in Kelvin so we don't have to worry about that.

Equals. We don't know what P₂ is. V₂ is 0.330 liters and then temperature 2 is 770 Kelvin. We're just looking to isolate our P₂ so you can cross multiply these and then you can cross multiply these so that you can isolate P₂. So if I come over here, I'm going to see that

P 2 = P 1 × 0.330   liters × 520   Kelvin = 1.85   atmospheres × 0.900   liters × 770   Kelvin

So divide out the volume and the temperature that we have here on both sides here. So we divide them both to both sides. By that we'll have isolated at the end our P₂, so liters cancel out, Kelvin's cancel out and we'll see here that P₂ = 7.47 atmospheres. We can see that by changing our volume and changing our temperature, it's caused this change in our pressure too, which comes out again to 7.47 atmospheres.