What would be the volume of hydrogen(2g/mol) needed in a balloon to lift a 50kg man? Assume that the temperature is 300K pressure is 1atm and air has average mass of 29 g/mol and that both gases behave ideally.
To find the volume of hydrogen needed to lift the 50kg man, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to calculate the number of moles of air in the atmosphere that the balloon has to lift. We can use the given average molar mass of air (29 g/mol) to find the number of moles:
n_air = mass_air / molar_mass_air
where:
mass_air = 50 kg (mass of the man)
molar_mass_air = 29 g/mol (average mass of air)
Converting the mass of the man from kg to grams:
mass_air = 50 kg * 1000 g/kg = 50000 g
Substituting these values, we have:
n_air = 50000 g / 29 g/mol ≈ 1724.14 mol
Now, we need to calculate the number of moles of hydrogen needed to lift the man. We can use the molar mass of hydrogen (2 g/mol) to find the number of moles:
n_hydrogen = mass_hydrogen / molar_mass_hydrogen
where:
mass_hydrogen = ?
molar_mass_hydrogen = 2 g/mol
Since we want to find the volume of hydrogen, we need to rearrange the ideal gas law equation to solve for V:
V = n_hydrogen * R * T / P
Substituting the given values into the equation:
P = 1 atm
T = 300 K
R = 0.0821 L·atm/mol·K
Now, to solve for mass_hydrogen, we can rearrange the equation:
mass_hydrogen = n_hydrogen * molar_mass_hydrogen
Substituting the values:
mass_hydrogen = n_air * molar_mass_hydrogen = 1724.14 mol * 2 g/mol = 34.482 g
Finally, substituting the values of mass_hydrogen, R, T, and P into the equation for V:
V = (34.482 g) * (0.0821 L·atm/mol·K) * (300 K) / (1 atm)
V ≈ 863.06 L
Therefore, approximately 863.06 liters of hydrogen would be needed in the balloon to lift the 50kg man.