Question

four 0.5 mol samples of krypton gas are described in the table below. rank these samples in order of decreasing rate of collisions between atoms and the walls of the container.
that is, select \1\ next to the sample with the most collisions per second between kr atoms and the walls of the container. select \2\ next to the sample with the next most collisions per second between kr atoms and the walls of the container, and so on.

Explanation:

Step1: Convert temperatures to Kelvin

• For sample A: \(T_A=-71^{\circ}C + 273 = 202\ K\)
• For samples B, C, D: \(T = - 54^{\circ}C+273 = 219\ K\)

Since temperature is a measure of the average kinetic energy of gas molecules, higher - temperature samples (all except A) have higher - speed molecules.

Step2: Analyze the effect of volume (for same - temperature samples B, C, D)

The ideal gas law \(PV = nRT\) can be related to the number of collisions. For a given \(n\) (number of moles) and \(T\) (temperature), \(P=\frac{nRT}{V}\). A smaller volume means a higher pressure. Pressure is related to the number of collisions of gas molecules with the walls of the container.

• For sample C (\(V = 28\ L\)), \(P_C=\frac{nRT}{28}\)
• For sample D (\(V = 32\ L\)), \(P_D=\frac{nRT}{32}\)
• For sample B (\(V = 36\ L\)), \(P_B=\frac{nRT}{36}\)

Since \(28<32<36\), \(P_C > P_D>P_B\) (for \(n\) and \(T\) constant)

Step3: Compare sample A with others

Since \(T_A(202\ K)<T_{B,C,D}(219\ K)\), sample A has the lowest - speed molecules among all.

Answer:

• Sample C: \(1\)
• Sample D: \(2\)
• Sample B: \(3\)
• Sample A: \(4\)