3 units edited Jul 4 '17 at 16:48 orthocresol♦ 43.4k77 gold badges136136 silver badges267267 bronze badges 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$$$m = \frac{\pu{101325 Pa} \times \pu{500 m3} \times \pu{28.8 g mol-1}}{\pu{8.31 J K-1 mol-1} \times \pu{290 K}^{[1]}}$$ where you can choose an appropriate value for [1] for your location. The rest of theAll values is(except for the molar mass) are in SI base units, as described here. 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$ where you can choose an appropriate value for [1] for your location. The rest of the values is in SI units, as described here. 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. 2 added 39 characters in body; added 78 characters in body edited Jul 4 '17 at 16:42 Glorfindel 1,53944 gold badges1111 silver badges2020 bronze badges 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$ where you can choose an appropriate value for [1] for your location. The rest of the values is in SI units, as described here. 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$ where you can choose an appropriate value for [1] for your location. 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$ where you can choose an appropriate value for [1] for your location. The rest of the values is in SI units, as described here. 1 answered Jul 3 '17 at 19:32 Glorfindel 1,53944 gold badges1111 silver badges2020 bronze badges 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: $$m = \frac{101325 \times 500 \times 28.8}{8.31 \times 290^{[1]}}$$ where you can choose an appropriate value for [1] for your location.