1. Given a gas at constant volume, pressure and temperature are related by Gay-Lussac's Law: $$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$ 2. From the first heating step, initial temperature $T_1 = 150$ K, final temperature $T_2 = 200$ K, and final pressure $p_2 = 110000$ Pa. 3. Calculate the initial pressure $p_1$: $$p_1 = p_2 \times \frac{T_1}{T_2} = 110000 \times \frac{150}{200} = 82500\ \text{Pa}$$ 4. Next, when the temperature increases further to 250 K, calculate final pressures for both cases: - For the red rectangle (initial conditions $p_1 = 82500$ Pa, $T_1 = 150$ K, $T_2 = 250$ K): $$p_2 = p_1 \times \frac{T_2}{T_1} = 82500 \times \frac{250}{150} = 137500\ \text{Pa}$$ - For the blue rectangle (initial conditions $p_1 = 110000$ Pa, $T_1 = 200$ K, $T_2 = 250$ K): $$p_2 = 110000 \times \frac{250}{200} = 137500\ \text{Pa}$$ Final answers rounded to three significant figures: - Initial pressure $p_1 = 82500$ Pa - Final pressure for both rectangles $p_2 = 138000$ Pa