hw2 - Homework 2 Solutions Spring 2010 Required Problems...

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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. 18-10b. 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. 18-10b 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. 19-8. ( 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.

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hw2 - Homework 2 Solutions Spring 2010 Required Problems...

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