A hot air balloon has a $\pu{500 m^3}$ volume. How much mass of normal air fits into the balloon, if the relative molar mass is $\pu{28.8 g/mol}$?

I am not sure if I should simply use the $PV=nRT$ formula and therefore
$$1\times500= m/28.8 \times 273 \times 8.31 .$$

I feel that there is something wrong, can anyone enlighten me with the right formula or the basic concept of solving this kind of questions?


You're right in using the ideal gas law, but the numbers are wrong. Assuming atmospheric conditions, the pressure is 101,325 Pascal and the temperature is likely to be higher than 0°C (273 K). So the mass of the gas (in grams) would be:

$$m = \frac{\pu{101325 Pa} \times \pu{500 m3} \times \pu{28.8 g mol-1}}{\pu{8.31 J K-1 mol-1} \times \pu{290 K}^{[1]}}$$

where you can choose an appropriate value for [1] for your location. All values (except for the molar mass) are in SI base units, as described here.


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