Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 18 - Exercises and Problems - Page 333: 79


17 percent

Work Step by Step

We know that 1100 tonnes of CO2 occupies $611,820 \ m^3$. Thus, using the equation for an adiabatic process, we find the final volume: $(1)(611,820)^{1.3}=350V^{1.3}\\ V = 6755 \ m^3$ Thus, we find the work needed using the equation for work for an adiabatic processes: $W = \frac{(6755)(350\times101,325)-(611,820)(101,325)}{1.3-1}=5.919 \times10^{11}\ J $ We divide this by the amount of energy produced in one hour: $ = \frac{5.919 \times10^{11}}{3600\times10^9}\approx 17 $%
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