Question

If 1 mole of H2 gas was collected under 294.2K and 746.7 mmHg, what volume will it occupy?

Answer 1

Well, `V=(nRT)/P`, and of course, we must choose the appropriate gas constant; `R=0.0821*L*atm*K^-1*mol^-1` is used here.

Explanation:

The key to solving these problems is to recall that `1*atm` of pressure will support a column of mercury that is `760*mm` high. A column of mercury may thus be used to measure pressures up to `1*atm`.

And so `V=(1*molxx0.0821*(L*atm)/(K*mol)xx294.2*K)/((746.7*mm*Hg)/(760*mm*Hg*atm^-1)`

`=??L`. It should be about `25*L`.