HW10-solutions

# HW10-solutions - baum(lmb2768 – HW10 – distler...

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Unformatted text preview: baum (lmb2768) – HW10 – distler – (56295) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Boltzmann’s constant is 1 . 38066 × 10- 23 J / K. Avogadro’s number is 6 . 02214 × 10 23 / mol. Determine the temperature at which the rms speed of an He atom equals 430 m / s. Correct answer: 29 . 651 K. Explanation: Let : v rms = 430 m / s , N A = 6 . 02214 × 10 23 / mol and k B = 1 . 38066 × 10- 23 J / K . Because the molecular mass of He is 4 g / mol, m He = 4 g / mol N A v rms = 3 k B T m He T = m He v rms 2 3 k B = m mol v rms 2 3 k B N A = (4 g / mol) (430 m / s) 2 3 (1 . 38066 × 10- 23 J / K) × 1 6 . 02214 × 10 23 / mol 1 kg 1000 g = 29 . 651 K . 002 (part 2 of 2) 10.0 points What is the rms speed of He on the surface of a certain star, where the temperature is 6809 K? Correct answer: 6516 . 14 m / s. Explanation: Let : T = 6809 K . v rms = 3 k B T m He = 3 k B T N A m mol = 3 (1 . 38066 × 10- 23 J / K) (6809 K) 4 g / mol × 6 . 02214 × 10 23 / mol 1000 g 1 kg = 6516 . 14 m / s . 003 10.0 points A sample of an ideal gas is in a tank of con- stant volume. The sample absorbs heat en- ergy so that its temperature changes from 214 K to 856 K. If v 1 is the average speed of the gas molecules before the absorption of heat and v 2 their average speed after the absorption of heat, what is the ratio v 2 v 1 ? 1. v 2 v 1 = 16 2. v 2 v 1 = 1 4 3. v 2 v 1 = √ 4 correct 4. v 2 v 1 = 1 5. v 2 v 1 = 4 Explanation: Let : T 1 = 214 K and T 2 = 856 K . For an ideal gas, the internal energy, (and therefore average kinetic energy of the molecules as well), is proportional to the tem- perature of the gas, i.e., U ∝ nRT K av ∝ k T ....
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HW10-solutions - baum(lmb2768 – HW10 – distler...

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