Chemistry - Ideal Gas Question with Solution | TestHub
ChemistryIdeal GasKinetic Theory of GasesMedium2 minQB
ChemistryMediumsingle choice
The RMS speed of a sample of ozone is 'v' m/s at 27°C. When the temperature is increased to 327°C, all the ozone molecules are dissociated into oxygen molecules. What is the RMS speed of new gas sample?
Options:
Answer:
D
Solution:
v = √(3R(300)/48) (O3, M = 48). New state: O2, M = 32, T = 327 + 273 = 600 K, so v' = √(3R(600)/32). Ratio: (v')/v = √((600/32)/)300/48 = √((600 × 48)/)32 × 300 = √(3). Hence v' = √(3) v. Confirms Ans (d).
Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Kinetic Theory of Gases
⏱ 2mℹ️ Source: QB
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