Exam 1 Answers

Exam 1 Answers - v rms = √ 3 RT = 459ms-1 see Table 2.4 1...

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NAME: EXAM I, ATMO 335, Feb 11, 2008. OPEN BOOK 1. A parcel of dry air has mass 2.8 kg, temperature 245 K. The parcel rests in the atmosphere at a height corresponding to a pressure of 500 hPa a) What is the volume of this parcel? V = M RT p = 3 . 94 m 3 b) What is the internal energy of this parcel? U = M c v T = 492 k J c) One gram of water is evaporated into this parcel. How much thermal energy is removed as the parcel cools? Q = M water L = 0 . 001 × 2 . 5 × 10 6 =2500 J d) How much does the internal energy change during this cooling? Δ T = Q M air c p = 0.89 K; Δ U = M air c v Δ T = 1787 J 2. Recall properties of the parcel in Problem 1 ( T = 245 K, p = 500 hPa, M = 2 . 8 kg). a) What is the number density n 0 in the parcel? n 0 = p k B T = 50000 1 . 38 × 10 - 23 × 245 = 1 . 48 × 10 25 molecules m - 3 b) What is the mean free path in the parcel (use a reasonable value of σ c )? Take σ c =0.5 nm 2 =0.5 × 10 - 18 m 2 . Then λ = 1 n 0 σ c 135 nm c) What is the rms speed of a molecule in the parcel?

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Unformatted text preview: v rms = √ 3 RT = 459ms-1 , see Table 2.4 1 d) What is the speed of sound in this parcel? v sound = √ γRT = 0 . 683 v rms = 314ms-1 ( γ = c p c v = 7 5 ) 3. A parcel of dry air is expanded adiabatically from 2.3m 3 to 5.4m 3 . Its initial pressure is p = 500hPa, and initial temperature T = 245K. a) What is the mass M of the parcel? M = pV RT = 50000 × 2 . 3 287 × 245 =1.64kg b) What is the ﬁnal pressure? p = p ( V V ) γ = 50000 ( 2 . 3 5 . 4 ) 1 . 400 = 151 . 4hPa c) What is the ﬁnal Temperature? T = T ± p p ² κ = 245 × ( 151 500 ) . 286 =174.0K d) What is the change in enthalpy? Δ H = M c p Δ T = 1 . 64 × 1005 × (174-245) =-117kJ 2...
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