Assignment 6 Solution

Assignment 6 Solution - meters deep, composed of 50% ice...

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PHYS-232 Heat and Waves - Homework 6 Winter Semester 2008 Due on ??, 2008 1. Solve problem 4.2.1 of the class note. Answer: ( P + a v 2 )( v - b ) = RT P = RT v - b - a v 2 P = nRT V - nb - an 2 V 2 (1) ( ∂P ∂V ) T = 2 n 2 a V 3 - nRT ( V - nb ) 2 (2) κ = - 1 V ( ∂V ∂P ) T = - 1 V [ nRT ( V - nb ) 2 - 2 an 2 V 3 ] - 1 = [ nRTV ( V - nb ) 2 - 2 an 2 V 2 ] - 1 (3) When a - > 0 , b - > 0: κ - > [ nRT V ] - 1 = P - 1 . 2. In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. ( γ air = 1 . 4) (a) What is the Fnal volume of this air after compression? (b) How much work is done in compressing the air? (c) If the temperature of the air is initially 300 K, what is the temperature after compression? Answer: a. Adiabatic process: V γ - 1 T = const, and V γ P = const. P increase from 1 atm to 7 atm: V γ f = 1 7 V γ i , V f = (1 / 7) 1 V i = 0 . 25 V i = 0 . 25 liter. b. W = i V f V i PdV = i V f V i C/V γ dV = C V 1 - γ 1 - γ | V f V i = P f V f - P i V i 1 - γ = 1 - 0 . 25 / 7 0 . 4 P i V i . Therefore the total work done is 2 . 41 × 101325pascals × 0 . 001m 3 = 244 . 3J. c. V γ - 1 T = const, V γ - 1 f T f = V γ - 1 i T i , therefore T f = (7) γ - 1 γ T i = 523 K. 3. When spring Fnally arrives in the mountains, the snow pack may be two
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Unformatted text preview: meters deep, composed of 50% ice and 50% air. Direct sunlight provides about 1000watts / m 2 to earths surface, but the snow might reect 90% of this energy. Estimate how many weeks the snow pack should last, if direct solar radiation is the only source of energy. (the latent heat for melting ice is 334 J/g ). Answer: The density of ice is 9 . 167 10 5 g / m 3 . So per square meter area there are 9 . 167 10 5 g / m 3 2m 3 50% = 9 . 167 10 5 g ice. It would take 9 . 167 10 5 g 334J / g / (1000Watts 10%) = 3061778s 35days 5weeks to melt. 4. Solve problem 4.8.1 of the class note. Answer: C = Q-W = T h T h-T c Therefore W / (1kJ) = 40 / 313 . 15, W = 127 . 7J. 1...
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