Exam 1 Answers

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
NAME: EXAM I, ATMO 335, Feb 11, 2008. OPEN BOOK 1. A parcel of dry air has mass 2.8kg, temperature 245K. The parcel rests in the atmosphere at a height corresponding to a pressure of 500hPa a) What is the volume of this parcel? V = M RT p = 3 . 94m 3 b) What is the internal energy of this parcel? U = M c v T = 492kJ 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 =2500J d) How much does the internal energy change during this cooling? Δ T = Q M air c p = 0.89K; Δ U = M air c v Δ T = 1787J 2. Recall properties of the parcel in Problem 1 ( T = 245K, p = 500hPa, M = 2 . 8kg). 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 moleculesm - 3 b) What is the mean free path in the parcel (use a reasonable value of σ c )? Take σ c =0.5nm 2 =0.5 × 10 - 18 m 2 . Then λ = 1 n 0 σ c 135nm c) What is the rms speed of a molecule in the parcel?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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 = RT = 0 . 683 v rms = 314ms-1 ( = 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 nal pressure? p = p ( V V ) = 50000 ( 2 . 3 5 . 4 ) 1 . 400 = 151 . 4hPa c) What is the nal Temperature? T = T p p = 245 ( 151 500 ) . 286 =174.0K d) What is the change in enthalpy? H = M c p T = 1 . 64 1005 (174-245) =-117kJ 2...
View Full Document

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online