ch18-p096 - the same value for path 1 as for path 2) then...

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96. Since the combination “ p 1 V 1 ” appears frequently in this derivation we denote it as “ x . Thus for process 1, the heat transferred is Q 1 = 5 x = Δ E int 1 + W 1 , and for path 2 (which consists of two steps, one at constant volume followed by an expansion accompanied by a linear pressure decrease) it is Q 2 = 5.5 x = Δ E int 2 + W 2 . If we subtract these two expressions and make use of the fact that internal energy is state function (and thus has
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Unformatted text preview: the same value for path 1 as for path 2) then we have 5.5 x 5 x = W 2 W 1 = area inside the triangle = 1 2 (2 V 1 )( p 2 p 1 ) . Thus, dividing both sides by x (= p 1 V 1 ), we find 0.5 = p 2 p 1 1 which leads immediately to the result: p 2 / p 1 = 1.5 ....
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