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 \(60 .^{\circ} \mathrm{C}\).

Understanding the pressure-volume relationship is crucial when studying the behavior of gases. According to Boyle's Law, for a fixed amount of gas at a constant temperature, pressure is inversely proportional to volume. This is often represented by the equation:
\[ P_1V_1 = P_2V_2 \]
where \(P_1\) and \(P_2\) are the initial and final pressures of the gas, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively. This means that if we decrease the volume of a gas by compressing it, the pressure will increase, provided that the temperature remains unchanged. Conversely, if we increase the volume, the pressure will decrease. This principle is crucial for explaining why option C from the original exercise would not double the pressure, as the presence of mercury changes the volume available to the gas but not the inherent pressure of the gas itself.

Gas moles calculation is fundamental in chemistry, especially when dealing with the ideal gas law. The amount of substance present in a system is measured in moles. One mole is equivalent to Avogadro's number \( (6.022 \times 10^{23}) \) of particles, whether they are atoms, molecules, ions, or electrons.
To calculate the number of moles \(n\) in a given mass \(m\) of a substance, we use the substance's molar mass \(M\) with the relationship:
\[ n = \frac{m}{M} \]
For instance, in the original exercise, the molar mass of nitrogen (N) is 28 g/mole, and 28 g of nitrogen is present. This results in:
\[ n = \frac{28 \, \text{g}}{28 \, \text{g/mole}} = 1 \, \text{mole} \]
This calculation demonstrates how the number of moles increases when additional gas is added. In option D, adding 32 g of oxygen, also equal to one mole (given that the molar mass of oxygen O is 32 g/mole), results in a total of two moles. This doubling of the mole count leads to doubling the pressure, assuming temperature and volume remain constant, which is in accordance with the ideal gas law.

The temperature-pressure relationship in gases is described by Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when volume is held constant. The mathematical expression of this law is a part of the ideal gas law and can be written as:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where \(P_1\) and \(P_2\) are the initial and final pressures, and \(T_1\) and \(T_2\) are the initial and final absolute temperatures in Kelvin, respectively. When solving problems involving temperature changes, it's essential to convert Celsius to Kelvin by adding 273.15 to the Celsius temperature.
In the given exercise, neither option B (raising temperature from -73°C to 127°C) nor E (raising temperature from 30°C to 60°C) resulted in the doubling of pressure because the change in temperature was not sufficient enough to cause a doubling in pressure. It's important to keep in mind that simply raising the temperature doesn't always lead to a precise proportional increase in pressure. The exercise improvements indicated that understanding this proportional relationship is key to correctly determining the outcomes in temperature-related scenarios.