## Essential University Physics: Volume 1 (3rd Edition)

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: $= \frac{5.919 \times10^{11}}{3600\times10^9}\approx 17$%