Assumptions:
* We'll assume standard temperature and pressure (STP) which is 0°C (273.15 K) and 1 atmosphere of pressure.
* We'll use the average molar mass of air, which is approximately 28.97 g/mol.
Helium
* Molar mass of helium (He): 4.003 g/mol
* Ideal Gas Law: PV = nRT (where P = pressure, V = volume, n = number of moles, R = ideal gas constant, T = temperature)
* Solving for n (number of moles): n = PV/RT
* Plugging in values: n = (1 atm)(1 L) / (0.0821 L*atm/mol*K)(273.15 K) = 0.0446 mol
* Mass of helium: mass = n * molar mass = (0.0446 mol) * (4.003 g/mol) = 0.179 grams
Air
* Molar mass of air: 28.97 g/mol
* Using the same Ideal Gas Law and calculations as above, we find the number of moles of air to be 0.0446 mol.
* Mass of air: mass = n * molar mass = (0.0446 mol) * (28.97 g/mol) = 1.29 grams
Therefore:
* The mass of 1 liter of helium at STP is approximately 0.179 grams.
* The mass of 1 liter of air at STP is approximately 1.29 grams.
