Question

How do you calculate the number of mol of a gas that is present in a 5.67 L container at 44.7 C, if the gas exerts a pressure of 614.0 mmHg.?

Answer 1

`n=(PV)/(RT)` `~=0.17*mol`

Explanation:

`n=(PV)/(RT)` `=` `(((614*mm*Hg)/(760*mm*Hg*atm^-1)xx5.67*L))/(0.0821*L*atm*K^-1*mol^-1xx317.9*K)`

`~=0.17*mol`

This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant, `R`. The most convenient unit of pressure is still the `"atmosphere"`, and this is a unit that is very intuitive. Given this, most chemists would use `mm*Hg`, knowing that `760*mm*Hg-=1*atm`, or rather that `1` `atm` of pressure will support a column of mercury that is `760*mm` high. And thus conversion of `mm*Hg` to `"atmospheres"` makes the choice of gas constant easy.