Question

What pressure (in atm) will 0.44 moles of `CO_2` exert in a 2.6 L container at 25°C?

Answer 1

`P=(nRT)/V=??*atm`

Explanation:

`P=(0.44*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)cancel(mol^-1)xx298*cancelK)/(2.6*cancelL)`

`=` `?? atm`.

The `"Gas constant"`, `R`, most generally used by chemists is `0.0821*L*atm*K^-1*mol^-1`. Sometimes, we use `dm^3`, i.e. `1*dm^3=(1xx10^-1*m)^3=10^-3*m^3`, i.e. `1L=1*dm^3=10^-3m^3`.