In order for a gas-filled balloon to rise in air, the density of the gas in the balloon must be less than that of air. Consider air at 25 degrees C and 1 atm to have a molar mass of 28.96 g/mol.
Determine the minimum temperature to which the balloon filled with argon at 1 atm would have to be heated before it could begin to rise in air. (Ignore the mass of the balloon itself.)
I have no idea how to go about solving the problem...If you could you list step by step how you got the answer that would be awesome. Thanks. :)
Determine the minimum temperature to which the balloon filled with argon at 1 atm would have to be heated before it could begin to rise in air. (Ignore the mass of the balloon itself.)
I have no idea how to go about solving the problem...If you could you list step by step how you got the answer that would be awesome. Thanks. :)
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Take a liter of air and apply the ideal gas law:
PV = nRT
n = PV/RT = (1 atm) x (1 L) / ((0.08205746 L atm/ K mol) x (25 + 273.15 K)) = 0.040874 mol
(0.040874 mol) x (28.96 g/mol) = 1.184 g
So the density of air is 1.184 g/L.
Now the question becomes: "At what temperature will Ar also have the density 1.184 g/L"?
(1.184 g/L) / (39.9481 g Ar/mol) = 0.02964 mol Ar/L
Solve the ideal gas law for T:
T = PV/nR
Rewrite it to include a (n/V) term (which is equal to the 0.02964 mol Ar/L found above):
T = (P/R) / (n/V) = ((1 atm) / (0.08205746 L atm/ K mol)) / (0.02964 mol Ar/L) = 411 K = 138 °C
PV = nRT
n = PV/RT = (1 atm) x (1 L) / ((0.08205746 L atm/ K mol) x (25 + 273.15 K)) = 0.040874 mol
(0.040874 mol) x (28.96 g/mol) = 1.184 g
So the density of air is 1.184 g/L.
Now the question becomes: "At what temperature will Ar also have the density 1.184 g/L"?
(1.184 g/L) / (39.9481 g Ar/mol) = 0.02964 mol Ar/L
Solve the ideal gas law for T:
T = PV/nR
Rewrite it to include a (n/V) term (which is equal to the 0.02964 mol Ar/L found above):
T = (P/R) / (n/V) = ((1 atm) / (0.08205746 L atm/ K mol)) / (0.02964 mol Ar/L) = 411 K = 138 °C