A hot-air balloon has a volume of 4.000 x 10^5 L at 30 degrees celsius and a pressure of 748 mmHg. If the average molar mass of air is 29.0 g/mol, what is the mass of air in the balloon?
Use PV = nRT and solve for n = number of moles. Then n = grams/molar mass and solve for grams.
Don't forget T must be in kelvin and P should be in atmospheres.
To find the mass of air in the balloon, we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L.atm/(mol.K))
T = Temperature (in Kelvin)
First, let's convert the given pressure from mmHg to atm:
1 atm = 760 mmHg
So, the pressure in atm is:
748 mmHg x (1 atm / 760 mmHg) = 0.9832 atm
Next, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
Temperature in Kelvin = 30°C + 273.15 = 303.15 K
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
n = (0.9832 atm) x (4.000 x 10^5 L) / [(0.0821 L.atm/mol.K) x (303.15 K)]
Calculating n:
n = 51.68 mol
Finally, to find the mass of air, we'll use the molar mass of air:
Mass = number of moles x molar mass
Mass = 51.68 mol x 29.0 g/mol
Calculating the mass:
Mass = 1499.12 g
Therefore, the mass of air in the balloon is approximately 1499.12 grams.