Question

gas a
the particles in gas a and gas b are moving at the same average speed. each particle in gas
b has more mass than each particle in gas a.
in which gas do the particles have the higher average kinetic energy?
gas a
gas b
neither; their particles have the
same average kinetic energy
which has the higher gas pressure?
gas a
gas b
neither; their gas pressures are
the same

Explanation:

Step1: Recall the formula for kinetic energy

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\), where \(m\) is mass and \(v\) is speed.

Step2: Compare kinetic energies of gas A and gas B

Given that \(v_{A}=v_{B}\) (average speeds are equal) and \(m_{B}>m_{A}\) (mass of particles in gas B is greater than in gas A).
Substitute into the kinetic - energy formula: \(K_{A}=\frac{1}{2}m_{A}v_{A}^{2}\) and \(K_{B}=\frac{1}{2}m_{B}v_{B}^{2}\). Since \(v_{A} = v_{B}\), and \(m_{B}>m_{A}\), we have \(K_{B}>K_{A}\) (because \(\frac{1}{2}v^{2}\) is a positive constant factor and \(m_{B}>m_{A}\)).

For gas pressure, pressure \(P\) in a gas (assuming same volume and number of particles) is related to the momentum transfer of particles. The momentum of a particle is \(p = mv\). When particles collide with the walls of the container, the change in momentum per unit time (which contributes to pressure) is considered. Since \(v_{A}=v_{B}\) and \(m_{B}>m_{A}\), the momentum of particles in gas B is greater. When they collide with the walls (assuming elastic collisions and same number of particles per unit volume), the force exerted on the walls (and thus pressure, \(P=\frac{F}{A}\)) is greater for gas B.

Answer:

gas B (for average kinetic energy)
gas B (for gas pressure)