# Ideal Gas Law: Where did R come from?

so what about our sometimes when people are learning the

## ideal

gas law they ask okay look here's this number I know that I can get it out of a textbook or off Wikipedia or whatever but## where

did the number actually## come

from I'm going to show you that right now so let it take the## ideal

gas law PV equals NRT and I'm going to rearrange it to solve for R so we're going to get R equals PV divided by and T so in order sulphur are we need to know the pressure volume number of moles andtemperature of a particular sample of gas well you know we could measure them and we could do that with a gas tank or something but there's a shortcut remember Avogadro's law Avogadro's law we use when we're doing moles and gas and it says that at STP which is one point zero zero atm and zero degrees Celsius one mole of any gas takes up twenty 2.4 liters so let's take the

## ideal

gas law rearranged here and just plug the values from Avogadro's law into it right so we canget our just assume that we're at STP and we have twenty 2.4 liters of gas all right so at STP we have one point zero zero ATM times a volume of 20 2.4 liters and one mole of gas is going to take up that much space and it's zero degrees Celsius we always want to use Kelvin we add 273 to the degrees Celsius so it's going to be 273 Kelvin okay so when we do that math the number that we get is zero point zero eight two zero five one now we go through three significant figures in each of

these so the guys the front are not significant but this one is this one is and this one is we looked next door to the Phi we round up or keep it the same we round up and check out what we get 0.0821 now we didn't cancel out any of our units here so the units for the final answer are going to be the exact same ones that they were in the in the original problem so it's going to be this atm times liters divided by moles times Kelvin it matches perfectly now the other ours we can get in the

exact same way remember these conversion factors for gases so what I'm going to do is I can set up this R equals PV divided by n T and I can solve for example for kPa instead of one atmosphere I'm going to put in 101.3 kPa check it out I do this math and I get 8.31 or I could take millimeters of mercury 760 plug that in instead of 1 atm I go through and I get sixty two point four millimeters of mercury so that quickly shows you the math that you can do to find out

## where

our came from thefirst place