23 - roofner (bar784) Extra Credit Homework 23 Weathers...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: roofner (bar784) Extra Credit Homework 23 Weathers (17104) 1 This print-out should have 15 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 Given: R = 8 . 31451 J / K mol . How much work is done by the steam when . 9 mol of water at 132 C boils and becomes . 9 mol of steam at 132 C at 1 . 6 atm pres- sure? Correct answer: 3030 . 37 J. Explanation: Given : n = 0 . 9 mol , R = 8 . 31451 J / K mol , T = 132 C = 405 K , m = 18 g / mol , P = 1 . 6 atm and = 1 g / cm 3 . Using the ideal gas law, P V = nRT . m is the molar mass of water. The work done is W = P V = nRT n parenleftbigg mP parenrightbigg = (0 . 9 mol) (8 . 31451 J / K mol) (405 K) (0 . 9 mol) (18 g / mol) (1 . 6 atm) 1 g / cm 3 parenleftBig m 100 cm parenrightBig 3 1 . 013 10 5 Pa 1 atm = 3030 . 37 J . 002 (part 2 of 2) 10.0 points Find the change in internal energy of the steam as it vaporizes. Consider the steam to be an ideal gas with heat of vaporization 2 . 26 10 6 J / kg. Correct answer: 33 . 5816 kJ. Explanation: Given : L v = 2 . 26 10 6 J / kg The heat needed to vaporize the water is Q = nmL v From the first law of thermodynamics U = Q W = nmL v W = bracketleftbigg (0 . 9 mol) (18 g / mol) 1 kg 1000 g (2 . 26 10 6 J / kg) 3030 . 37 J bracketrightbigg 1 kJ 1000 J = 33 . 5816 kJ . 003 10.0 points One mole of an ideal gas is heated at constant pressure so that its temperature increases by a factor of 9. Then the gas is heated at constant temperature so that its volume increases by a factor of 9. Find the ratio of the work done during the isothermal process to that done during the isobaric process. Correct answer: 2 . 47188. Explanation: Given : k = 9 . The work done in the isobaric process is W ab = integraldisplay b a P dV = P a ( V b V a ) = nR ( k T a ) nRT a = ( k 1) nRT a , The work done in the isothermal process is W bc = integraldisplay c b P dV = nRT b integraldisplay c b dV V = nRT b ln parenleftbigg V c V b parenrightbigg = nR ( k T a ) ln k , Therefore the ratio fraction is r = W bc W ab = nR ( k T a ) ln k ( k 1) nRT a = k ln k k 1 = 9 ln(9) (9 1) = 2 . 47188 . roofner (bar784) Extra Credit Homework 23 Weathers (17104) 2 004 10.0 points The surface of the Sun has a temperature of about 6214 K.about 6214 K....
View Full Document

This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.

Page1 / 5

23 - roofner (bar784) Extra Credit Homework 23 Weathers...

This preview shows document pages 1 - 3. Sign up to view the full document.

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