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?


1 Answer 1


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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