Answered: In the fermentation of glucose (wine…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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In the fermentation of glucose (wine making), a volume of 780 mL carbon dioxide gas was
produced at 37 oC and 1.00 atm. What is the volume (L) of the gas when measured at 22 oC and
675 mm Hg?

Expert Solution

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$G i v e n : V_{1} = 780 m L = 0.78 L V_{2} = ? T_{1} = (273.15 + 37) K = 310.15 K T_{2} = (273.15 + 22) K = 295.15 K P_{1} = 1.00 a t m P_{2} = 675 m m H g = 0.89 a t m A c c o r d i n g t o i d e a l g a s l a w . (\frac{P_{1} V_{1}}{T_{1}}) = (\frac{P_{2} V_{2}}{T_{2}}) . . . . . . . . . . . . . . . . . . 1 s t S u b s t i t u t i n g t h e v a l u e s i n e q u a t i o n 1 s t . (\frac{1.00 a t m \times 0.78 L}{310.15 K}) = (\frac{0.89 a t m \times V_{2}}{295.15 K}) V_{2} = (\frac{1.00 a t m \times 0.78 L \times 295.15 K}{0.89 a t m \times 310.15 K}) V_{2} = (\frac{230.217}{276.0335}) L V_{2} = 0.8340 L = 834 m L (answer)$

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