Question

the diagrams below show three pure samples of gas in identical closed, rigid containers. each sample contains the same number of gas particles. the particles in gas a and gas c are moving at the same average speed, while the particles in gas b are moving at a higher average speed. each particle in gas a and gas b has the same mass, while each particle in gas c has less mass. order the samples by their gas pressures. higher gas pressure gas c gas b gas a lower gas pressure

Explanation:

Step1: Analyze the effect of particle speed on pressure

According to the kinetic - molecular theory of gases, pressure is related to the force and frequency of particle collisions with the container walls. Faster - moving particles (higher average speed) will collide with the walls more forcefully and more frequently. Since gas B has particles moving at a higher average speed than gas A (and gas C), gas B will have a higher pressure than gas A (and initially higher than gas C based on speed alone).

Step2: Analyze the effect of particle mass on pressure

The momentum of a particle is \(p = mv\). For gas A and gas C, \(v_{A}=v_{C}\), but \(m_{A}>m_{C}\). The force exerted during a collision is related to the change in momentum. When a particle collides with the wall, \(\Delta p = 2mv\) (assuming an elastic collision and normal incidence for simplicity). Since \(m_{A}>m_{C}\) and \(v_{A} = v_{C}\), the change in momentum (and thus the force) for gas A particles is greater than for gas C particles. Also, the frequency of collisions is the same for gas A and gas C (same number of particles in the same - sized container). So, \(P=\frac{F}{A}\), and since \(F_{A}>F_{C}\) (where \(F\) is the total force due to particle - wall collisions and \(A\) is the area of the container walls), \(P_{A}>P_{C}\)

Answer:

gas B, gas A, gas C