Chemistry - Ideal Gas Question with Solution | TestHub
ChemistryIdeal GasIdeal Gas Equation & problemsEasy2 minQB
ChemistryEasyinteger
What will be the temperature difference (in °C) needed in a hot air balloon to lift 1.91 kg of mass? Assume that the volume of the balloon is 100 m³, the temperature of the ambient air is 27°C, the pressure is 1.0 bar, and air is an ideal gas with an average molar mass of 29 g/mole. (R = 0.08314 L-bar/K-mol)
Answer:
5
Solution:
Ambient air in 100 m³ (10⁵ L) at 300 K, 1.0 bar: n = PV/RT = 10⁵/(0.08314 × 300) = 4009 mol; mass = 4009 × 29 = 116.27 kg.
Required hot-air mass = 116.27 - 1.91 = 114.36 kg, i.e. n = 114360/29 = 3943 mol.
Tₕot = PV/nR = 10⁵/(3943 × 0.08314) = 305 K = 32 °C.
Delta T = 32 - 27 = 5 °C.
Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems
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