A closed auditorium of volume \(2.50 \cdot 10^{4} \mathrm{~m}^{3}\) is filled with 2000 people at the beginning of a show, and the air in the space is at a temperature of \(293 \mathrm{~K}\) and a pressure of \(1.013 \cdot 10^{5} \mathrm{~Pa}\). If there were no ventilation, by how much would the temperature of the air rise during the \(2.00-\mathrm{h}\) show if each person metabolizes at a rate of \(70.0 \mathrm{~W} ?\)

To determine the total heat produced during the show, we need to multiply the metabolic rate of each person by the number of people and the duration of the show in seconds. \(Q = \text{Number of people} \times \text{Metabolic rate} \times \text{Duration of the show in seconds}\) \(Q = 2000 \times 70.0 \, \text{W} \times 2 \, \text{h} \times 3600 \, \frac{\text{s}}{\text{h}}\) \(Q = 1004 \times 10^6 \, \text{J}\)

Using the ideal gas law, we can determine the number of moles of air inside the auditorium. \(PV = nRT\) We will rearrange the formula to find n: \(n = \frac{PV}{RT}\) We are given the initial pressure (\(P = 1.013 \times 10^5 \, \text{Pa}\)), volume (\(V = 2.50 \times 10^4 \, \text{m}^3\)), and temperature (\(T = 293 \, \text{K}\)) of the air, and the gas constant (\(R = 8.314 \, \frac{\text{J}}{\text{mol}\cdot\text{K}}\)). \(n = \frac{(1.013 \times 10^5 \, \text{Pa})(2.50 \times 10^4 \, \text{m}^3)}{(8.314 \, \frac{\text{J}}{\text{mol} \cdot \text{K}})(293 \, \text{K})}\) \(n \approx 1.03 \times 10^6 \, \text{mol}\)

We will now apply the first law of thermodynamics, which states: \(\Delta U = Q - W\) In this case, there is no work done on or by the system, so \(W = 0\). The change in internal energy is given by: \(\Delta U = n \, C_v \, \Delta T\) Where \(C_v\) is the molar heat capacity at constant volume for air, which is approximately \(20.8 \, \frac{\text{J}}{\text{mol} \cdot \text{K}}\). We can solve for the change in temperature: \(Q = n \, C_v \, \Delta T\) \(\Delta T = \frac{Q}{n \, C_v}\) \(\Delta T = \frac{1004 \times 10^6 \, \text{J}}{(1.03 \times 10^6 \, \text{mol})(20.8 \, \frac{\text{J}}{\text{mol} \cdot \text{K}})}\) \(\Delta T \approx 4.7 \, \text{K}\) The temperature of the air would rise by approximately \(4.7 \, \text{K}\) during the show without ventilation.