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?

Explanation:

Step1: Calculate the average kinetic energy of Sample A

The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\). For Sample A, \(m = 46u\) and \(v=1400m/s\). So, \(KE_{A}=\frac{1}{2}\times46\times(1400)^{2}=4.508\times10^{7}u\cdot m^{2}/s^{2}\)

Step2: Calculate the average kinetic energy of Sample B

For Sample B, \(m = 34u\) and \(v = 1300m/s\). So, \(KE_{B}=\frac{1}{2}\times34\times(1300)^{2}=2.873\times10^{7}u\cdot m^{2}/s^{2}\)

Step3: Compare the average kinetic energies

Since \(KE_{A}>KE_{B}\), and temperature is a measure of the average kinetic energy of particles in a substance.

Answer:

Sample A