Ideal Gas Law: Where did R come from?

So what about our times, when people are learning the

ideal

gas law, you ask? Well, look, here's this number. I know I can get it from a textbook or Wikipedia or whatever, but

where

did the number actually

come

from? I'm going to show it. that right now, so let's take the

ideal

gas law: PV equals NRT and I'm going to rearrange it to solve for R, so we'll get R equals PV divided by and T, so to find out if there is sulfur the pressure , the volume, the number of moles and the temperature of a particular sample of gas, well, you know, we could measure them and we could do it with a tank of gasoline or something, but there is a shortcut, remember Avogadro's law.

Avogadro's law that we use when we do moles and gas and it says that at STP, which is zero point zero atm and zero degrees Celsius, one mole of any gas occupies twenty 2.4 liters, so let's take the ideal gas law rearranged here and let's just plug Avogadro's law values ​​into it correctly so that We can assume that we are at STP and we have twenty 2.4 liters of gas, so at STP we have a zero point zero ATM multiplied by a volume of 20 2 .4 liters and one mole of gas will consume that amount. In space and it's zero degrees Celsius, we always want to use Kelvin, we add 273 to the degrees Celsius, so it will be 273 Kelvin.

Well, when we do those calculations, the number we get is zero point zero eight two zero five one. We are going now. through three significant figures in each of these, so the guys in front are not significant, but this is this one and this one we look next to the Phi, we round or keep it the same, we round and check whatever you get 0.0821 now no we cancel out any of our units here, so the units for the final answer will be exactly the same as they were in the original problem, so it will be this atm multiplied by liters divided by moles multiplied by Kelvin matches perfectly now with our other one we can get exactly the same way.

Remember these conversion factors for gases, so what I'm going to do is set this R equal to PV divided by n T and I can solve it, for example. for kPa instead of an atmosphere, I'm going to put 101.3 kPa, check it out. I do these calculations and I get 8.31 or I could take 760 millimeters of mercury, plug that instead of 1 atm, I step up and get sixty-two points. four millimeters of mercury to quickly show you the calculations you can do to find out

where

the first one came from