Question

Given that the average speed of an \(\mathrm{O}_{2}\) molecule is 1,700 \(\mathrm{km} / \mathrm{h}\) at \(0^{\circ} \mathrm{C}\), what would you expect the average speed of a \(\mathrm{CO}_{2}\) molecule to be at the same temperature?

Short answer

The average speed of a CO₂ molecule at the same temperature as an O₂ molecule with a speed of 1,700 km/h would be approximately 396.19 m/s.

Step-by-step solution

Identify relevant data and formula

We are given: - the average speed of an O₂ molecule (v₁) = 1,700 km/h - corresponding molecular mass of O₂ (M₁) = 32 g/mol - molecular mass of CO₂ (M₂) = 44 g/mol We need to find the average speed of a CO₂ molecule (v₂) at the same temperature. The formula that relates the average speeds of two molecules of different masses at the same temperature comes from the concept of root-mean-square speed: \(v_{1} \propto \sqrt{\frac{1}{M_{1}}}\) \(v_{2} \propto \sqrt{\frac{1}{M_{2}}}\) Since both speeds are proportional to the same temperature, we can write the relationship: \(\frac{v_{1}}{v_{2}} =\sqrt{\frac{M_{2}}{M_{1}}}\)

Convert speed to the same unit

First, we need to convert the given average speed of O₂ from km/h to m/s. To do this, divide the given speed by 3.6: \(v_{1} = \frac{1700 \,\text{km/h}}{3.6} = 472.22 \, \text{m/s}\)

Calculate the average speed of CO₂

Now, we can use the formula derived in step 1 and substitute the values: \(\frac{472.22}{v_{2}} =\sqrt{\frac{44}{32}}\) Solve for \(v_2\): \(v_{2} = \frac{472.22}{\sqrt{\frac{44}{32}}}\) After solving, we get the average speed of CO₂ at the same temperature: \(v_{2} \approx 396.19 \,\text{m/s}\) Therefore, the average speed of a CO₂ molecule at the same temperature is approximately 396.19 m/s.

Key concepts

Root-Mean-Square Speed

The term "root-mean-square speed" is crucial when discussing molecular speed in gases. It represents the speed of gas particles when temperature and molecular mass come into play. Essentially, root-mean-square speed (often abbreviated as RMS speed) is a type of average. It focuses on the mathematical mean of the squares of the individual speeds of molecules. More formally, it's derived from the equation:\[ v_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}} \]Here's what the components mean:

• \( k \) is the Boltzmann constant, a fundamental physical constant.
• \( T \) signifies the absolute temperature, measured in Kelvin.
• \( R \) is the universal gas constant.
• \( m \) is the mass of a single molecule.
• \( M \) stands for the molar mass of the gas.

You can think of root-mean-square speed as a way to understand how fast, on average, the molecules in a gas are moving at a given temperature and with a specific molecular mass. It gives a more accurate picture than simply using arithmetic averages and is deeply rooted in kinetic molecular theory.

Molecular Mass

Molecular mass is a key player in determining the speed of gas molecules. It refers to the mass of a given molecule and is often measured in atomic mass units (u) or grams per mole (g/mol). Essentially, a higher molecular mass means that a molecule will generally move slower than one with a lower mass when at the same temperature.When comparing molecules like \( \text{O}_2 \) and \( \text{CO}_2 \), each with different molecular masses, the one with the larger mass (\( \text{CO}_2 \) at 44 g/mol) will typically move at a slower speed than a smaller mass molecule (\( \text{O}_2 \) at 32 g/mol).In gas laws and kinetic theory:

• Molecular speeds are inversely proportional to the square root of their masses.
• This relationship can be represented as a mathematical expression: \(v \propto \sqrt{\frac{1}{M}}\).
• This implies that when you compare two gases, the lighter one will move faster.

Understanding this helps explain why lighter gases, like helium, escape the Earth's atmosphere more easily than heavier gases.

Temperature Dependence

Temperature has a significant impact on the speed of gas molecules. It's directly related to the kinetic energy of particles. As the temperature increases, the energy available to the molecules increases, causing them to move more quickly.In the context of root-mean-square speed:

• The speed of gas particles is proportional to the square root of the temperature in Kelvin: \( v_{rms} \propto \sqrt{T} \).
• Higher temperatures lead to higher speeds because the particles have more energy to overcome intermolecular forces.
• At absolute zero, a theoretical temperature, molecular motion would cease entirely because kinetic energy would be nonexistent.

Thus, when analyzing problems involving gases, assuming constant temperature helps simplify solutions. It allows the use of proportional relationships between molecular speed and temperature without the need for extra adjustments for varying thermal conditions. This principle is critical in understanding and predicting the behavior of gas molecules under different thermal environments.