Question

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

Short answer

The correct process that will double the pressure exerted on the walls of the glass vessel containing 28 g of nitrogen gas, assuming ideal behavior, is \(d. \text{ Adding 28 g nitrogen gas}\).

Step-by-step solution

Recall the Ideal Gas Law

The Ideal Gas Law is given by the equation PV = nRT, where P represents pressure, V is the volume of the gas, n is the amount of gas in moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Convert the mass of nitrogen gas to moles.

To use the Ideal Gas Law, we first need to convert the mass of nitrogen gas (28 g) to the amount of the substance in moles. The molar mass of nitrogen gas (N₂) is 28 g/mol. So, moles of nitrogen gas (n) = \(\frac{28 \mathrm{~g}}{28 \mathrm{~g/mol}}\) = 1 mol.

Analyze each option

We can now analyze each option given in the problem statement. a. Adding half the volume of mercury to the container only changes the volume of the container and not the volume of the gas, and therefore it does not affect the pressure of the gas. b. Raising the temperature from 30°C to 60°C (303.15 K to 333.15 K) will cause a change in pressure but not necessarily a doubling of pressure. c. Raising the temperature from -73°C to 127°C (-200.15 K to 400.15 K) also results in a change in pressure but again, not necessarily a doubling of pressure. d. Adding 28 g of nitrogen gas (another 1 mol) doubles the amount of nitrogen gas present in the container. Assuming the volume remains constant, doubling the number of moles of nitrogen gas will double the pressure.

Select the correct option

Based on the analysis of each option and the Ideal Gas Law, the correct process that will double the pressure exerted on the walls of the vessel is: d. Adding 28 g nitrogen gas.

Key concepts

Gas Pressure

Gas pressure is a fundamental concept in the study of gases, and understanding it is crucial for grasping the behavior of gases under different conditions. In a practical sense, gas pressure is the force that a gas exerts on the walls of its container. This force is the result of countless collisions of gas particles with the container's surfaces.

When considering the ideal gas law, which is summarized by the equation \( PV = nRT \), pressure (\( P \)) is directly proportional to the number of moles of gas (\( n \)) and the temperature (\( T \)), while inversely proportional to the volume (\( V \)). So, when the amount of gas in a container is increased without changing its volume, the pressure will rise accordingly, as there will be more particles colliding with the container's walls more frequently.

• In scenario (d) from the textbook exercise, by adding another 28 grams of nitrogen gas, we double the number of gas particles in the container while keeping the volume constant. The result is a doubling of the pressure, which is directly linked to the increase in the number of moles as reflected in the ideal gas law.

Molar Mass of Nitrogen

The molar mass of a substance is the mass of one mole of that substance. In the case of nitrogen gas (\( N_2 \)), which is composed of two nitrogen atoms, the molar mass is crucial for converting between the mass of the gas and the amount in moles. This step ensures that calculations using the ideal gas law are accurate.

Nitrogen gas has a molar mass of 28 grams per mole (g/mol). Knowing this allows us to calculate the number of moles from the given mass in the exercise. For instance, with 28 grams of nitrogen gas, we have precisely one mole (\( \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \text{ mol} \)). This quantity is directly plugged into the ideal gas law to determine how changes in conditions affect the pressure.

• By understanding the molar mass of nitrogen, we can manipulate the number of moles to predict how the pressure within a container will alter, as shown in the proposed solutions to the textbook exercise.

Temperature and Pressure Relationship

The relationship between temperature and pressure for a gas is governed by Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume is kept constant. In practice, this means that if you increase the temperature of a gas, its pressure will increase, assuming the gas's volume does not change, a concept which is integral to the ideal gas law.

In the exercise, scenarios (b) and (c) reflect changes in temperature. However, without proper calculations, it's not immediately clear if these changes would result in doubling the pressure. Only exact mathematical relationships like the one used to solve for (d) can confirm the outcome:

For (b):

• The temperature change from 30°C to 60°C is significant but without arithmetic, we can't confirm that pressure would double.

For (c):

• The temperature change from -73°C to 127°C is even more dramatic, yet again, we'd need to use the ideal gas law formula to see if it precisely doubles the pressure.

Thus, a thorough understanding of how temperature affects pressure is key to solving these types of problems correctly and is exemplified by the solution to the textbook exercise.