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kinetic energy of ideal gas

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A 1.1mol sample of hydrogen gas has a temperature of 24 degree C. What is the total kinetic energy of all the gas molecules in the sample? Anw: 6800J How fast would a 65-{\rm kg} person have to run to have the same kinetic energy? ANW: 14m/s

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A 1.1mol sample of hydrogen gas has a temperature of 24 degree C. What is the total kinetic of all the gas molecules in the sample? (The temperature should be 224°C instead of 24°C . Otherwise, the answer would be 4100 J.) Number of moles, n = 1.1 mol Absolute temperature, T = (273 + 224) K = 497 K Gas constant = 8.314 J/(mol K) Kinetic energy, Ek = (3/2)nRT = (3/2) x 1.1 x 8.314 x 497 J = 6800 J (2 sig. fig.) How fast would a 65-kg person have to run to have the same kinetic energy? Ek = 6800 J m = 65 kg Ek = (1/2)mv^2 v^2 = 2Ek/m v = √(2Ek/m) Speed, v = √(2 x 6800/65) m/s = 14 m/s (2 sig. fig.)

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