P21_027 - Q out = nC p T = 5 2 ( p a V a p c V c ) = 5 2 p...

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27. (a) Energy is added as heat during the portion of the process from a to b . This portion occurs at constant volume ( V b ), so Q in = nC V T . The gas is a monatomic ideal gas, so C V = 3 2 R and the ideal gas law gives ∆ T =(1 /nR )( p b V b p a V a )=(1 /nR )( p b p a ) V b .Thu s , Q in = 3 2 ( p b p a ) V b . V b and p b are given. We need to Fnd p a .Now p a isthesameas p c and points c and b are connected by an adiabatic process. Thus, p c V γ c = p b V γ b and p a = p c = µ V b V c γ p b = µ 1 8 . 00 5 / 3 (1 . 013 × 10 6 Pa) = 3 . 167 × 10 4 Pa . The energy added as heat is Q in = 3 2 (1 . 013 × 10 6 Pa 3 . 167 × 10 4 Pa)(1 . 00 × 10 3 m 3 )=1 . 47 × 10 3 J . (b) Energy leaves the gas as heat during the portion of the process from c to a
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Unformatted text preview: Q out = nC p T = 5 2 ( p a V a p c V c ) = 5 2 p a ( V a V c ) = 5 2 (3 . 167 10 4 Pa)( 7 . 00)(1 . 00 10 3 m 3 ) = 5 . 54 10 2 J . The substitutions V a V c = V a 8 . 00 V a = 7 . 00 V a and C p = 5 2 R were made. (c) or a complete cycle, the change in the internal energy is zero and W = Q = 1 . 47 10 3 J 5 . 54 10 2 J = 9 . 18 10 2 J. (d) The eciency is = W/Q in = (9 . 18 10 2 J) / (1 . 47 10 3 J) = 0 . 624....
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