91Ó°ÊÓ

Using the Ideal Gas Law, \(PV = nRT\), we can calculate the number of moles \(n\) of nitrogen gas inside the balloon. Firstly, we need to convert the temperature from Celsius to Kelvin, \(T_i = 20.0^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 293.15 \mathrm{K}\) and \(T_f = 40.0^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 313.15 \mathrm{K}\). Given \(V = 200.0 \mathrm{L}\), \(T_i = 293.15 \mathrm{K}\), and \(P = 1 \mathrm{atm}\). Then we have \(1 \mathrm{atm} \cdot 200.0 \mathrm{L} = n \cdot 0.0821 \frac{\mathrm{L.atm}}{\mathrm{K.mol}} \cdot 293.15 \mathrm{K}\) Rearranging and solving for \(n\) gives: $$n = \frac{1\mathrm{atm} \cdot 200.0\mathrm{L}}{0.0821 \frac{\mathrm{L.atm}}{\mathrm{K.mol}} \cdot 293.15\mathrm{K}}$$ $$n \approx 8.17 \mathrm{mol}$$

Now we will use the energy equation \(q = n C_p \Delta T\) to calculate the amount of energy required to raise the temperature from \(T_i = 293.15\mathrm{K}\) to \(T_f = 313.15\mathrm{K}\), while keeping the pressure constant. Recall that for diatomic gases, the specific heat capacity under constant pressure is \(C_p = \frac{7}{2} R \approx 29.1 \frac{\mathrm{J}}{\mathrm{mol.K}}\). The temperature difference is \(\Delta T = T_f - T_i = 313.15\mathrm{K} - 293.15\mathrm{K} = 20.0\mathrm{K}\). Using the energy equation with \(n \approx 8.17 \mathrm{mol}\), \(C_p \approx 29.1 \frac{\mathrm{J}}{\mathrm{mol.K}}\), and \(\Delta T = 20.0\mathrm{K}\), we get: $$q = 8.17 \mathrm{mol} \cdot 29.1 \frac{\mathrm{J}}{\mathrm{mol.K}} \cdot 20.0\mathrm{K}$$ $$q \approx 4760 \mathrm{J}$$ So, approximately 4760 J of energy must be added to raise the temperature of the nitrogen to \(40.0^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure.