What mass (in gram of air) is filled in a hot balloon when it is finflated (at 95oC) to a volume of 500 L at 1.15 atm? (The average molar mass of air is 29 g/mol)
Usw PV = nRT and solve for n = number of mols air.
Then n = grams/molar mass. Solve for grams. Don't forget T must be in kelvin.
To find the mass of air filled in the hot balloon, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 95 + 273.15 = 368.15 K
Now we can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substituting the given values:
n = (1.15 atm) * (500 L) / [(0.0821 L·atm/(mol·K)) * (368.15 K)]
Calculating:
n ≈ 1.15 * 500 / (0.0821 * 368.15)
n ≈ 70.14 mol
Next, we can calculate the mass of air:
mass = n * molar mass
Substituting the given average molar mass of air:
mass = 70.14 mol * 29 g/mol
mass ≈ 2034.06 g
Therefore, approximately 2034.06 grams of air is filled in the hot balloon when it is inflated to a volume of 500 L at 95°C and 1.15 atm.