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A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double its r.m.s. speed of its molecules will be; Take R = 8.32 J mol⁻¹ K⁻¹.
- 2229 J
- 5616 J
- 9360 J
- 13,104 J
Correct answer: 9360 J
Solution
14 g N2 = 0.5 mol at 300 K. Doubling rms speed makes T = 4×300 = 1200 K, so ΔT = 900 K. With diatomic Cv = 5R/2, Q = 0.5 × (5/2)(8.32) × 900 = 9360 J.
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