problem20_48

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

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20.48: For a monatomic ideal gas, . C and 2 3 2 5 R R C V P = = a) ab : The temperature changes by the same factor as the volume, and so J. 10 25 . 2 ) m Pa)(0.300 10 00 . 3 )( 5 . 2 ( ) ( 5 3 5 × = × = - = = b a a P P V V p R C T nC Q The work V p is the same except for the factor of J. 10 90 . 0 so , 5 2 5 × = W J. 10 1.35 5 × = - = W Q U bc : The temperature now changes in proportion to the pressure change, and J, 10 40 . 2 ) m Pa)(0.800 10 00 . 2 )( 5 . 1 ( ) ( 5 3 5 2 3 × - = × - = - = b b c V p p Q and the work is zero J. 10 40 . 2 ). 0 ( 5 × - = - = = W Q U V ca: The easiest way to do this is to find the work done first; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is
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Unformatted text preview: × × + × Pa) 10 1.00 Pa 10 00 . 3 ( 5 5 2 1 J, 10 00 . 6 ) m 500 . m (0.800 4 3 3 × =-and so the work is be must J. 10 60 . 5 U ∆ ×-(since J 10 1.05 5 = ∆ × U for the cycle, anticipating part (b)), and so Q must be J. 10 45 . 5 × = + ∆ W U b) See above; . J, 10 30 . 5 = ∆ × = = U W Q c) The heat added, during process ab and ca, is 2.25 J 10 0.45 J 10 5 5 × + × J 10 2.70 5 × = and the efficiency is %. 1 . 11 111 . 5 5 H 10 70 . 2 10 30 . = = = × × Q W...
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