Question

At \(27^{\circ} \mathrm{C}\) the ratio of root mean square speeds of ozone to oxygen is (1) \(\sqrt{\frac{3}{5}}\) (2) \(\sqrt{\frac{4}{3}}\) (3) \(\sqrt{\frac{2}{3}}\) (4) \(0.25\)

Short answer

The ratio is \( \sqrt{\frac{2}{3}} \) which corresponds to option (3).

Step-by-step solution

Recall the formula for root mean square speed

The root mean square speed (\text{v}_{\text{rms}}) of a gas is given by the formula: \[ v_{rms} = \sqrt{\frac{3kT}{M}} \]where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass (in kg/mol).

Compare root mean square speeds of ozone and oxygen

For two gases, the ratio of their root mean square speeds is given by: \[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{M_2}{M_1}} \]Here, gas 1 is ozone (\( O_3 \)) and gas 2 is oxygen (\( O_2 \)).

Determine molar masses

The molar mass of ozone (\( O_3 \)) is \( 3 \times 16 = 48 \) g/mol, and the molar mass of oxygen (\( O_2 \)) is \( 2 \times 16 = 32 \) g/mol.

Substitute molar masses into the ratio formula

Using the molar masses, we substitute into the ratio formula: \[ \frac{v_{rms(O_3)}}{v_{rms(O_2)}} = \sqrt{\frac{32}{48}} = \sqrt{\frac{2}{3}} \]

Match the answer

The ratio \( \sqrt{\frac{2}{3}} \) corresponds to the option given in the problem as (3).

Key concepts

kinetic theory of gases

The kinetic theory of gases helps us understand how gases behave based on the idea that they are made up of tiny, moving particles. This theory explains several properties of gases:

• Particles are constantly moving in random directions.
• Collisions between gas particles and the walls of their container cause pressure.
• The average kinetic energy of gas particles is directly proportional to the temperature in Kelvin.

This theory is crucial for understanding why different gases at the same temperature can have different speeds. It's all about the energy and movement of the gas particles. A related concept we often use is the root mean square speed (\text{v}_{\text{rms}}), which is a measure of the average speed of particles in a gas. The formula is \text{v}_{\text{rms}} = \sqrt{\frac{3kT}{M}}, where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

molar mass comparison

Molar mass is the mass of one mole of a substance and it affects the speed of gas molecules. By comparing the molar masses of different gases, we can learn about their relative speeds. For example:

• Ozone (\( O_{3} \)) has a molar mass of 48 g/mol.
• Oxygen (\( O_{2} \)) has a molar mass of 32 g/mol.

Given these values, we can use the formula for the ratio of root mean square speeds to determine how their speeds compare:
\( \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{M_{2}}{M_{1}}} \)
Substituting the molar masses of ozone and oxygen, we get:

\( \frac{v_{rms(O_{3})}}{v_{rms(O_{2})}} = \sqrt{\frac{32}{48}} = \sqrt{\frac{2}{3}}\)
This ratio means that ozone molecules move slower relative to oxygen molecules because ozone has a greater molar mass.

temperature and gas speed relationship

Temperature affects the speed of gas particles. According to the kinetic theory of gases, higher temperatures increase the kinetic energy of particles, making them move faster. This relationship is shown in the root mean square speed formula:
\[ v_{rms} = \sqrt{\frac{3kT}{M}} \]
Here, as \( T \) (the temperature) increases, \( v_{rms} \) (the speed) also increases, assuming \( M \) (the molar mass) stays constant. Let's break it down:

• If you heat a gas, the particles gain more energy and move faster.
• For equal temperatures, gases with heavier particles (higher molar mass) will have slower speeds compared to lighter ones.

In the example problem, we looked at gases at \( 27^{\circ} \mathrm{C} \) or 300 Kelvin. By knowing this, we can always compare how the speed of gas particles would change with temperature variations. Remember, when solving these kinds of problems, always convert the temperature to Kelvin for accurate calculations.