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Unformatted text preview: Homework 2 Solutions Spring 2010 Required Problems Chapter 18 48. The mean free path is given by Eq. 1810b. Combine this with the ideal gas law to find the mean free pathdiameter relationship. ( 29 ( 29 M 2 2 1 2 M 1 ; 4 2 4 2 2 N P kT PV NkT V kT r N V d P kT d P = = = = = l l ( a ) ( 29 ( 29 ( 29 ( 29 23 10 8 5 M 1.38 10 J K 273K 3.9 10 m 2 2 5.6 10 m 1.013 10 Pa kT d P  = = = l ( b ) ( 29 ( 29 ( 29 ( 29 23 10 8 5 M 1.38 10 J K 273K 1.8 10 m 2 2 25 10 m 1.013 10 Pa kT d P  = = = l 72. We combine the ideal gas law with Eq. 1810b for the mean free path. From problem 45, we se that the diameter of the average air molecule is 10 3 10 m. Since air is mostly nitrogen molecules, this is a good approximation for the size of a nitrogen molecule. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 M 2 2 2 23 8 M 2 2 10 5 ; 1 1 4 2 4 2 4 2 1.38 10 J K 300K 1.4 10 m 4 2 4 2 1.5 10 m 7.5 1.013 10 Pa N P PV NkT V kT kT r N V r P kT r P kT r P  = = = = = = = = l l Note that this is about 100 times the radius of the molecules. Chapter 19 25. The kinetic energy of the bullet is assumed to warm the bullet and melt it. ( 29 ( 29 [ ] ( 29 ( 29 ( 29 2 1 Pb melt initial fusion 2 5 Pb melt initial fusion 2 2 130J kg C 327 C 20 C 0.25 10 J kg 360m s mv Q mc T T mL v c T T L = = + = + =  + = g 33. ( a )The work done by an ideal gas during an isothermal volume change is given by Eq. 198. ( 29 ( 29 ( 29 3 2 3 1 7.00m ln 2.60mol 8.314J mol K 290K ln 4345.2J 4350J 3.50m V W nRT V = = = g ( b )Since the process is isothermal, there is no internal energy change. Apply the first law of thermodynamics. int 0 4350J E Q W Q W = = = = ( c ) Since the process is isothermal, there is no internal energy change, and so int E = . 34. ( a )Since the process is adiabatic, 0 J Q = ( b ) Use the first law of thermodynamics to find the change in internal energy. The work is done on the gas, and so is negative. ( 29 int 2850J 2850 J E Q W = =  = ( c ) Since the internal energy is proportional to the temperature, a rise in internal energy means a rise in temperature. 39. We are given that ac 85J, Q =  ac 55 J, W =  cda 38 J, W = int,a int,b int ba 15J, E E E = = and a d 2.2 . P P = ( a )Use the first law of thermodynamics to find int,a int,c ca . E E E = ( 29 ( 29 int int ac ac ca ac 85J 55J 30J E E Q W =  =  =    = ( b )Use the first law of thermodynamics to find cda Q . int cda cda cda int cda int cda cda cda ca 30 J 38 J 68J E Q W Q E W E W = = + = + = + = ( c ) Since the work along path bc is 0, ( 29 abc ab a ab a b a W W P V P V V = = = ....
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This note was uploaded on 04/23/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at University of California, Berkeley.
 Spring '08
 ZETTLE

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