HW2_solution

HW2_solution - PHYSICS 140A : ASSIGNMENT #2 SOLUTIONS...

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Unformatted text preview: PHYSICS 140A : ASSIGNMENT #2 SOLUTIONS Problem 1. Solution: (a) As we know B V and B T , we can directly use: Pa V RT P B B B 6 10 04 . 1 ! = = " (b) We can compute the area under the curve to get the work in this process. ) ( C B B B A V V P PdV PdV W ! ! = = " " ) 1 ( B C B VB VA A V V RT dV V RT ! ! = " # # J V V V V RT B C A B B 3 10 64 . 2 1 ln ! = " # \$ % & ’ + ( ) ) * + , ,- . = / (c) ! ! ! + = = = " C B C B C B T PdV dE T Q d dS S (1) where 1 ! = " PV E , take it into (1), we get: P dP R V dV R dV T P T VdP PdV S C B C B ! ! " + " = + " + = # 1 1 ) 1 ( \$ % % \$ \$ \$ K J V V R B C / 3 . 40 ln 1 ! = " " # \$ % % & ’ ! = ( ) ) Problem 2. Solution: First of all, let’s consider the work during this period, i.e. ABCD S ! ! ! ! + + + = = A D D C C B B A ABCD PdV PdV PdV PdV S W ! ! " + + = D C D A A D A B A A A V V P dV V V P dV V V P ) ( # # # # ) ( 1 1 ) ( 1 1 ! ! ! ! ! ! D A B A D A A V V V P V V P " " " " " = " Secondly, in this period, the ideal gas absorb heat during D->A, therefore:...
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This note was uploaded on 02/06/2009 for the course PHYSICS 1101 taught by Professor Hasan during the Spring '08 term at Temple.

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HW2_solution - PHYSICS 140A : ASSIGNMENT #2 SOLUTIONS...

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