Question

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

Short answer

The correct answer is option D: Adding 32 g of oxygen gas. This doubles the initial number of moles in the container while keeping temperature and volume constant, resulting in a doubled pressure.

Step-by-step solution

Option A: Adding 28 g of oxygen gas.

The number of moles of nitrogen gas, n, is given by: n = \( \frac{28}{28} = 1 \) mole After adding 28 g of oxygen gas, the total number of moles becomes: total n = \( \frac{28}{28} + \frac{28}{32} \) moles, where 28 g/mole is the molar mass of nitrogen gas and 32 g/mole is the molar mass of oxygen gas. It is clear that the total number of moles is increasing in this case. If we want to double the pressure, we also need to double the number of moles, keeping the temperature and volume constant. Therefore, option A is not correct.

Option B: Raising the temperature from -73°C to 127°C

We can use the ideal gas law to see the effect of temperature on pressure. Given that the initial pressure is P1 and the final pressure is P2: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) We are asked if the pressure doubles when the temperature increases from -73°C to 127°C. First, we need to convert these temperatures to Kelvin: T1 = -73 + 273.15 = 200.15 K T2 = 127 + 273.15 = 400.15 K Now we can check if the pressure doubles: \( \frac{P_1}{200.15} = \frac{2P_1}{400.15} \) Clearly, the pressure does not double with this temperature change. Option B is not correct.

Option C: Adding enough mercury to fill one-half the container

Adding mercury to the container will not affect the number of gas moles, temperature, or pressure directly. As mercury is liquid and does not mix with the nitrogen gas, this option will not double the pressure. Option C is not correct.

Option D: Adding 32 g of oxygen gas

Adding 32g of oxygen gas means the total number of moles in the container is: total n = \( \frac{28}{28} + \frac{32}{32} \) moles This results in a total of 2 moles, which is double the initial number of moles. Since temperature and volume are constant, the pressure will double in this case. Option D is correct.

Option E: Raising the temperature from 30°C to 60°C

As we did for option B, we can check if the pressure doubles when the temperature increases from 30°C to 60°C. We convert these temperatures to Kelvin: T1 = 30 + 273.15 = 303.15 K T2 = 60 + 273.15 = 333.15 K We can then check if the pressure doubles: \( \frac{P_1}{303.15} = \frac{2P_1}{333.15} \) In this case, the pressure does not double. Option E is not correct. So, the correct answer is option D: Adding 32 g of oxygen gas.

Key concepts

Pressure-Volume Relationship

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

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.

Temperature-Pressure Relationship

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.