P21_064 - constant-volume step). Using Eq. 20-14 and Eq....

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64. (a) The most obvious input-heat step is the constant-volume process. Since the gas is monatomic, we know from Chapter 20 that C V = 3 2 R . Therefore, Q V = nC V T =( 1m o l ) µ 3 2 ¶µ 8 . 31 J mol · K (600 K 300 K) = 3740 J . Since the heat transfer during the isothermal step is positive, we may consider it also to be an input-heat step. The isothermal Q is equal to the isothermal work (calculated in the next part) because ∆ E int = 0 for an ideal gas isothermal process (see Eq. 20-45). Borrowing from the part (b) computation, we have Q isotherm = nRT H ln 2 = (1 mol) µ 8 . 31 J mol · K (600 K) ln2 = 3456 J . Therefore, Q H = Q V + Q isotherm =7 . 2 × 10
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Unformatted text preview: constant-volume step). Using Eq. 20-14 and Eq. 20-16, we have W = nRT H ln V max V min + p min ( V min V max ) where (by the gas law in ratio form, as illustrated in Sample Problem 20-1) the volume ratio is V max V min = T H T L = 600 K 300 K = 2 . Thus, the net work is W = nRT H ln 2 + p min V min 1 V max V min = nRT H ln 2 + nRT L (1 2) = nR ( T H ln 2 T L ) = (1 mol) 8 . 31 J mol K ((600 K) ln2 (300 K)) = 9 . 6 10 2 J . (c) Eq. 21-9 gives = W Q H = 0 . 134 13% ....
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