A glass vessel contains \(28 \mathrm{~g}\) of nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding \(28 \mathrm{~g}\) of oxygen gas b. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\) c. Adding enough mercury to fill one-half the container d. Adding \(32 \mathrm{~g}\) of oxygen gas e. Raising the temperature of the container from \(30 .{ }^{\circ} \mathrm{C}\) to \(60 .{ }^{\circ} \mathrm{C}\)

Before analyzing each option, we must understand the relationships within the Ideal Gas Law. We know that the pressure exerted on the walls of the vessel is influenced by the volume, the number of moles of gas, the gas constant R, and the temperature T. Since the gas constant, R, is always the same, our main concern here is how the addition of gas, change in volume, or change in temperature will affect the pressure of the system.

Now that we understand how each component influences the pressure in the Ideal Gas Law, let's analyze each option. a. Adding 28 g of oxygen gas: This would increase the number of moles of gas in the container, and since the volume is constant, it can increase the pressure. We need to check if it doubles the pressure or not. b. Raising the temperature of the container from -73°C to 127°C: Since the Ideal Gas Law is directly proportional to temperature, increasing the temperature will also increase the pressure. We need to check if this increase in temperature doubles the pressure. c. Adding enough mercury to fill one-half the container: Adding mercury would decrease the volume available for the gas. This could potentially increase the pressure. We must check if this will double the pressure. d. Adding 32 g of oxygen gas: Similar to option a, we need to check if adding this extra oxygen gas will double the pressure or not. e. Raising the temperature of the container from 30°C to 60°C: Similar to option b, we must check if this increase in temperature doubles the pressure.

Let's now check each option - a. Adding 28 g of oxygen gas: To check this, we must look at the molecular weights of Nitrogen and Oxygen. Molecular weight of Nitrogen (N₂) = \(28 \mathrm{~g/mol}\) Molecular weight of Oxygen (O₂) = \(32 \mathrm{~g/mol}\) For Nitrogen, moles added = \(\frac{28}{28}\) = \(1 \mathrm{~mol}\) of nitrogen For Oxygen, moles added = \(\frac{28}{32}\) = \(0.875 \mathrm{~mol}\) of oxygen This process adds more moles of gas, but not enough to double the pressure. b. Raising the temperature of the container from -73°C to 127°C: To check this, we must convert the temperatures to Kelvin. Initial temperature: -73 + 273.15 = 200.15 K Final temperature: 127 + 273.15 = 400.15 K Doubling the pressure requires doubling the temperature in this instance since the volume and the number of moles of gas are constant. Since 400.15K is exactly double 200.15K, this process doubles the pressure. c. Adding mercury to fill one-half of the container: Since the flexible container walls are in contact with both the mercury and the gas, reducing the volume by half would double the pressure, so this process works. d. Adding 32 g of oxygen gas: For Oxygen, moles added = \(\frac{32}{32}\) = \(1 \mathrm{~mol}\) of oxygen In this process, we are adding exactly double the moles of the initial amount of nitrogen. Since the temperature and volume are constant, the pressure will also double. So this process works too. e. Raising the temperature of the container from 30°C to 60°C: Converting to Kelvin - Initial temperature: 30 + 273.15 = 303.15 K Final temperature: 60 + 273.15 = 333.15 K In this case, the temperature has not doubled and since other factors are constant we can confidently say that the pressure has not doubled either.

To summarize the findings, the processes that can double the pressure exerted on the walls of the vessel are: b. Raising the temperature of the container from -73°C to 127°C c. Adding enough mercury to fill one-half the container d. Adding 32 g of oxygen gas