Chemistry - Ideal Gas Question with Solution | TestHub
ChemistryIdeal GasIdeal Gas Equation & problemsMedium2 minQB
ChemistryMediumnumerical
What will be the temperature difference (in °C) needed in a hot air balloon to lift 10.0/8.314 kg of mass? Assume that the volume of the balloon is 91.0 m3, the temperature of the ambient air is 17°C, the pressure is 1.0 bar, and air is an ideal gas with an average molar mass of 29 g/mole.
Answer:
3.00
Solution:
Lift = (mass of displaced ambient air) - (mass of hot air inside) = 10.0/8.314 kg = 1.203 kg.
Ambient air in 91.0 m3 (91000 L) at 290 K, 1.0 bar: n = PV/RT = (1.0 × 91000)/(0.08314 × 290) = 3774 mol; mass = 3774 × 29 = 109.45 kg.
Hot air mass inside must be 109.45 - 1.20 = 108.25 kg, i.e. n = 108250/29 = 3733 mol.
Thot = PV/nR = (1.0 × 91000)/(3733 × 0.08314) = 293.2 K = 20.2 °C.
Delta T = 20.2 - 17.0 = 3 °C (nearest integer).
Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems
⏱ 2mℹ️ Source: QB
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