Question

the diagrams below show two pure samples of gas in identical closed, rigid containers. each colored ball represents one gas particle. both samples have the same number of particles.
compare the average kinetic energies of the particles in each sample. which sample has the higher temperature?
sample a
sample b

Explanation:

Step1: Recall the formula for kinetic energy

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

Step2: Compare the average kinetic energies

Since the mass \(m\) of each particle is the same (\(m = 16u\)) for both samples, the kinetic energy depends only on the speed \(v\).
For sample A, \(v_{A}=940m/s\), so \(KE_{A}=\frac{1}{2}m\times(940)^{2}\).
For sample B, \(v_{B}=930m/s\), so \(KE_{B}=\frac{1}{2}m\times(930)^{2}\).
Since \(940>930\), then \(KE_{A}>KE_{B}\).

Step3: Relate kinetic energy to temperature

According to the kinetic - molecular theory, the average kinetic energy of gas particles is directly proportional to the temperature (\(KE_{avg}\propto T\)).

Answer:

Sample A