You are not sure whether to fill a balloon with He or hot air. To what temperature would the air have to be heated for a balloon to rise to the same height has the balloon filled wit He at 25.0 C? Please show the steps you used to work this problem out.
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In order for the two balloons to behave in the same manner, the density of what fills them must be the same. So the problem becomes: At what temperature would air have the same density as He at 25.0 °C?
Supposing all this occurs at a pressure of 1 atm:
Calculate the density of 1.000 L of He at 25.0 °C and 1 atm:
(1.000 L) x (273/(25 + 273)) / (22.4 L/mol) x (4.0026 g He/mol) = 0.1637 grams He
So He at 25.0 °C has a density of 0.1637 g/L.
The effective molar mass of air is 28.97 g/mol according to the source below. (Since there is no mention of humidity in the problem, water vapor is neglected here -- although as a practical matter there will be lots of water vapor in a hot air balloon since the hot air comes from the combustion of hydrocarbons which produces water vapor.)
So making the same calculation for air, without knowing the temperature:
(1.000 L) x (273/T) / (22.4 L/mol) x (28.97 g/mol) = 0.1637 grams air
Solve for T algebraically:
T = 2157°K = 1884°C
This is absurdly hot and would ignite any hot air balloon's envelope. It also makes the concern for humidity in the air moot.
Supposing all this occurs at a pressure of 1 atm:
Calculate the density of 1.000 L of He at 25.0 °C and 1 atm:
(1.000 L) x (273/(25 + 273)) / (22.4 L/mol) x (4.0026 g He/mol) = 0.1637 grams He
So He at 25.0 °C has a density of 0.1637 g/L.
The effective molar mass of air is 28.97 g/mol according to the source below. (Since there is no mention of humidity in the problem, water vapor is neglected here -- although as a practical matter there will be lots of water vapor in a hot air balloon since the hot air comes from the combustion of hydrocarbons which produces water vapor.)
So making the same calculation for air, without knowing the temperature:
(1.000 L) x (273/T) / (22.4 L/mol) x (28.97 g/mol) = 0.1637 grams air
Solve for T algebraically:
T = 2157°K = 1884°C
This is absurdly hot and would ignite any hot air balloon's envelope. It also makes the concern for humidity in the air moot.