problem19_42

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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19.42: (a) abc ac v abc ac abc W T nC W U Q + = + = get nR PV T nRT PV T ac = = : nR V P V P nR V P nR V P T T T a a c c a a c c a c ac - = - = - = K 289 K) mole J 8.31 mole)( ( ) m Pa)(0.0020 10 0 . 1 ( ) m Pa)(0.010 10 0 . 1 ( 3 1 3 5 3 5 = × - × = ac T Pa) 10 0 . 1 ( m 0.002) (0.010 Pa) 10 5 . 2 ( m ) 002 . 0 010 . 0 ( 2 1 graph under Area 5 3 5 3 × - + × - = = PV W abc ( 29 gas the into J 3000 J 10 8 . 1 J 10 20 . 1 J 10 20 . 1 K 289 K mole J 31 . 8 2 3 mole 3 1 2 3 J 10 80 . 1 3 3 3 ac 3 = × + × = × = = = = × = abc ac ac v abc Q T R n T nC U W ac U (b) in the same = 1200 J
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Unformatted text preview: ( 29 ( 29 gas the into J 2000 J 800 J 1200 J 800 Pa 10 . 1 m 002 . 010 . area 5 3 = + = + ∆ = = ×-= = ac ac ac ac W U Q W (c) More heat is transfered in abc than in ac because more work is done in abc....
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• NoProfessor
• 1 mole, 10 pA, 8.31 J, 10 J, 2 J

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