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Chapter 5: Problem 20

You have two containers each with 1 mole of xenon gas at $15^{\circ} \mathrm{C}\( . Container A has a volume of \)3.0 \mathrm{L},$ and container \(\mathrm{B}\) has a volume of \(1.0 \mathrm{L}\) . Explain how the following quantities compare between the two containers. a. the average kinetic energy of the Xe atoms b. the force with which the Xe atoms collide with the container walls c. the root mean square velocity of the Xe atoms d. the collision frequency of the Xe atoms (with other atoms) e. the pressure of the Xe sample

Short Answer

Expert verified
In summary: a. The average kinetic energy of Xe atoms is the same in both containers, as they have the same temperature. b. The force with which Xe atoms collide with the container walls is greater in container B, as it has a higher pressure. c. The root mean square velocity of Xe atoms is the same in both containers, as they have the same temperature and gas type. d. The collision frequency of Xe atoms (with other atoms) is greater in container B, as it has a smaller volume. e. The pressure of the Xe sample is greater in container B, due to its smaller volume compared to container A.

Step by step solution

01

Find the temperature in Kelvin.

Convert the given temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15 T(K) = 15°C + 273.15 = 288.15 K
02

Calculate the pressure of each container using the Ideal Gas Law.

The ideal gas law is given by: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J/(mol K)), and T is the temperature in Kelvin. For container A: P_A = (1 mol)(8.314 J/(mol K))(288.15 K) / (3.0 L * 0.001 m^3/L) ≈ 7835 Pa For container B: P_B = (1 mol)(8.314 J/(mol K))(288.15 K) / (1.0 L * 0.001 m^3/L) ≈ 23505 Pa
03

Compare the average kinetic energy of Xe atoms in both containers.

The average kinetic energy of gas particles is given by: KE = (3/2)kT, where k is the Boltzmann constant (1.38 × 10⁻²³ J/K). Since both containers have the same temperature, the average kinetic energy will be the same for both containers.
04

Compare the force with which Xe atoms collide with container walls.

The force of particles colliding with the container walls is directly proportional to the gas pressure. Since P_B > P_A, the force exerted by the Xe atoms on the wall of container B is greater than that on the wall of container A.
05

Compare the root mean square velocity of Xe atoms in both containers.

Root mean square velocity is given by: v_rms = √(3kT/m), where m is the molar mass of the gas. Since both containers have the same temperature and gas, the root mean square velocity of Xe atoms will be the same in both containers.
06

Compare the collision frequency of Xe atoms (with other atoms) in both containers.

Collision frequency is inversely proportional to the square root of the volume. As the volume of container A is three times greater than the volume of container B, the collision frequency will be higher in container B.
07

Compare the pressure of the Xe sample in both containers.

We have already calculated the pressures in containers A and B in Step 2. The pressure of the Xe sample in container B is greater than that in container A. To summarize the results: a. The average kinetic energy of Xe atoms is the same in both containers. b. The force with which Xe atoms collide with the container walls is greater in container B. c. The root mean square velocity of Xe atoms is the same in both containers. d. The collision frequency of Xe atoms (with other atoms) is greater in container B. e. The pressure of the Xe sample is greater in container B.

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