Question

Consider separate \(1.0\) -L gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0\) -L samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

Short answer

a. At STP, all gases have the same average kinetic energy: KE_avg(H2) = KE_avg(Xe) = KE_avg(Cl2) = KE_avg(O2). b. The order of increasing average velocity is H2 < O2 < Cl2 < Xe based on their molar masses. c. For O2 and H2 to have the same average velocity, the temperature of O2 should be 16 times the temperature of H2.

Step-by-step solution

a. Rank the gases in order of increasing average kinetic energy

According to the kinetic molecular theory, the average kinetic energy of an ideal gas is directly proportional to its temperature: \(KE_{avg} = \cfrac{3}{2} kT\) where KE_avg is the average kinetic energy, k is the Boltzmann constant, and T is the temperature. Since the gases are all at STP (Standard Temperature and Pressure), their temperatures are the same (273.15 K). Therefore, their average kinetic energies are also equal. Hence, KE_avg(H2) = KE_avg(Xe) = KE_avg(Cl2) = KE_avg(O2)

b. Rank the gases in order of increasing average velocity

Average velocity of a gas can be related to its molar mass using the following equation derived from the kinetic molecular theory: \(v_{avg} = \sqrt{\cfrac{8RT}{\pi M}}\) where v_avg is the average velocity, R is the gas constant, T is the temperature, and M is the molar mass. Since the temperature is the same for all gases, we can rank the gases based on their molar mass (as lighter gases have higher average velocities): - H2: M = 2 g/mol - O2: M = 32 g/mol - Cl2: M = 70.9 g/mol - Xe: M = 131.3 g/mol Therefore, the order of increasing average velocity is: H2 < O2 < Cl2 < Xe

c. How can separate 1.0 L samples of O2 and H2 each have the same average velocity?

To have the same average velocity for O2 and H2 gases, their temperature and mass should be adjusted. Based on the average velocity equation given in part b, we can equate the two expressions for O2 and H2: \(\sqrt{\cfrac{8R*T_O}{\pi * M_O}} = \sqrt{\cfrac{8R*T_H}{\pi * M_H}}\) Where TO and TH are the temperatures of O2 and H2 respectively, and MO and MH are their molar masses. We can simplify the equation and solve for one of the temperatures: \(T_O = \cfrac{M_O*T_H}{M_H}\) Let's assume the mass and temperature of H2 are constant and find the corresponding temperature for O2. Using the molar masses from part b: \(T_O = \cfrac{32*T_H}{2}\) \(T_O = 16 * T_H\) This means that in order for O2 and H2 to have the same average velocity, the temperature of O2 should be 16 times the temperature of H2.

Key concepts

Average Kinetic Energy

When we talk about the average kinetic energy of gas molecules, we're referring to the energy they possess due to their motion. It's determined by the temperature of the gas and not by the type or size of the gas molecules.

Using the kinetic molecular theory, we learn that average kinetic energy (\(KE_{avg}\)) is given by \(KE_{avg} = \cfrac{3}{2} kT\) where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in kelvins (K).

If you have a group of gases at the same temperature, like at Standard Temperature and Pressure (STP), they all have the same average kinetic energy. This means that heavier gases don't move slower because they're heavier, and lighter gases don't move faster because they're lighter — they move at velocities that impart the same kinetic energy on average, purely because they're at the same temperature.

STP (Standard Temperature and Pressure)

STP stands for Standard Temperature and Pressure, which are conditions that set a baseline to compare different scientific experiments and calculations. At STP, the temperature is 0 degrees Celsius (273.15 K) and the pressure is 1 atmosphere (atm).

By using STP, we can ensure that variables like temperature and pressure don't influence the outcomes of our experiments when we want to focus on other factors. For instance, in the exercise, all gases are compared at STP to isolate the effects of molar mass on their average velocities. This helps in understanding the intrinsic properties of the gases without external factors skewing the results.

Molar Mass

Molar mass is the mass of one mole of a substance (usually in grams per mole). It is a physical property that is unique to each element or compound and can be thought of as the 'weight' of a molecule on a microscopic scale.

The molar mass plays a crucial role when examining the behavior of gases under the same temperatures and pressures. In the problem's solution, we observe that gases with lower molar masses will have higher average velocities. This is an insightful piece of information because it tells us that even though molecules with more mass have more inertia, they move at such speeds that their kinetic energies even out with lighter, faster-moving molecules of other gases at the same temperature.

Average Velocity of Gas

The average velocity of gas molecules is a measure of how fast the molecules move. The formula \(v_{avg} = \sqrt{\cfrac{8RT}{\pi M}}\) indicates that it is dependent on the temperature (T), the universal gas constant (R), and inversely dependent on the molar mass (M) of the gas.

So, at a constant temperature (as with STP), gases with a smaller molar mass move faster on average. This explains why in the textbook exercise, hydrogen (\(\text{H}_2\)), with the smallest molar mass, has the highest average velocity, despite all gases being at the same kinetic energy levels. It's a fine example of how molar mass can be a deciding factor in the behavior of gases when conditions like temperature are equalized.