# 5 a in one minute the person generates q 100 j s60 s

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number of conformations for the unfolded state. 5. (a) In one minute, the person generates q = ( - 100 J / s)(60 s) = - 6000 J of heat. The oxidation reaction is C 6 H 12 O 6(aq) + 6O 2(aq) 6CO 2(aq) + 6H 2 O (l) . The enthalpy of reaction is Δ r H m = 6Δ f H (CO 2 , aq) + 6Δ f H (H 2 O , l) - f H (C 6 H 12 O 6 , aq) + 6Δ f H (O 2 , aq)] = 6( - 413 . 26) + 6( - 285 . 830) - [ - 1263 . 06 + 6( - 12 . 09)] kJ / mol = - 2858 . 94 kJ / mol n C 6 H 12 O 6 = - 6 . 00kJ - 2858 . 94 kJ / mol = 0 . 002 10 mol . 1

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(b) n O 2 = 6 n C 6 H 12 O 6 = 0 . 0126 mol. (c) The number of moles of gas per breath is calculated from the ideal gas law: V = (0 . 5 L) / (1000 L / m 3 ) = 5 × 10 - 4 m 3 . n = pV RT = (101 325 Pa)(5 × 10 - 4 m 3 ) (8 . 314 472 J K - 1 mol - 1 )(293 . 15 K) = 0 . 0208 mol . Of this amount, 21% is oxygen, so the amount of oxygen in each breath is 4 . 36 × 10 - 3 mol. 25% of the latter amount is actually absorbed, so the amount of oxygen absorbed per breath is 1 . 09 × 10 - 3 mol / breath. To take in 0.0126 mol of oxygen in a minute, we must therefore take 0 . 0126 mol 1 . 09 × 10 - 3 mol / breath = 11 . 5 breaths per minute . (d) Work of expansion against a constant external pressure: w = - p ext Δ V . Each breath expands the chest by 0.5 L, so 11.5 breaths expands the chest by a total of 5.77 L, or 5 . 77 × 10 - 3 m 3 . Thus, w = - (101 325 Pa)(5 . 77 × 10 - 3 m 3 ) = - 585 J . 2
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