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Chapter 5: Problem 149

A glass vessel contains 28 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 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 g of oxygen gas e. Raising the temperature of the container from \(30 .^{\circ} \mathrm{C}\) to \(60 .^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Processes b, c, and d will double the pressure exerted on the walls of the vessel.

Step by step solution

01

Process a: Adding 28 g of oxygen gas

Since both nitrogen and oxygen are ideal gases, their pressures add up when mixed. First, we need to calculate the moles of each gas. For nitrogen, n = mass/molar mass = 28/28 = 1 mole, and for oxygen, n = mass/molar mass = 28/32 = 0.875 moles. The total number of moles is now 1 + 0.875 = 1.875 moles. Assuming volume and temperature remain constant, we can use the ideal gas law to calculate the ratio of the new pressure to the initial pressure: P_new/P_initial = n_new/n_initial = 1.875/1 = 1.875, which is not doubling the pressure.
02

Process b: Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\)

Changing the temperature affects the pressure directly, assuming volume and moles remain constant. First, let's convert the given temperatures to Kelvin: T_initial = (273 - 73) K = 200 K and T_final = (273 + 127) K = 400 K. Now we can use the ideal gas law to calculate the ratio of the new pressure to the initial pressure: P_new/P_initial = T_new/T_initial = 400/200 = 2, which results in doubling the pressure.
03

Process c: Adding enough mercury to fill one-half the container

Since the mercury occupies half the container, the volume of the gas decreases to half of the initial volume. Assuming temperature and moles of gas remain constant, we can use the ideal gas law to calculate the ratio of the new pressure to the initial pressure: P_new/P_initial = V_initial/V_new = 2/1 = 2, which doubles the pressure.
04

Process d: Adding 32 g of oxygen gas

First, we need to calculate the moles of oxygen gas added: n = mass/molar mass = 32/32 = 1 mole. The total number of moles is now 1 + 1 = 2 moles. Assuming volume and temperature remain constant, we can use the ideal gas law to calculate the ratio of the new pressure to the initial pressure: P_new/P_initial = n_new/n_initial = 2/1 = 2, which doubles the pressure.
05

Process e: Raising the temperature of the container from \(30 .^{\circ} \mathrm{C}\) to \(60 .^{\circ} \mathrm{C}\)

First, let's convert the given temperatures to Kelvin: T_initial = (273 + 30) K = 303 K and T_final = (273 + 60) K = 333 K. Assuming volume and moles remain constant, we can use the ideal gas law to calculate the ratio of the new pressure to the initial pressure: P_new/P_initial = T_new/T_initial = 333/303 ≈ 1.10, which is not doubling the pressure. In conclusion, processes b, c, and d will double the pressure exerted on the walls of the vessel.

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