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A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g/mol) is moving on a floor at a speed of 50 m/s. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.
3 years ago

Navjyot Kalra
654 Points
Sol. N = 1 mole, W = 20 g/mol, V = 50 m/s

K.E. of the vessel = Internal energy of the gas
= (1/2) mv^2 = (1/2) × 20 × 10^–3 × 50 × 50 = 25 J
25 = n 3/2 r (∆T) ⇒ 25 = 1 * 3/2 * 8.31 * ∆T ⇒ ∆T = 50/3 * 8.3 = 2 k.
3 years ago
Kevin Nash
332 Points
Sol. 1. N = 1 mole, W = 20 g/mol, V = 50 m/s

K.E. of the vessel = Internal energy of the gas
= (1/2) mv^2 = (1/2) × 20 × 10^–3 × 50 × 50 = 25 J
25 = n 3/2 r (∆T) ⇒ 25 = 1 * 3/2 * 8.31 * ∆T ⇒ ∆T = 50/3 * 8.3 = 2 k.
3 years ago
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