Question

The root mean square speed of hydrogen molecules at \(300 \mathrm{~K}\) is \(1930 \mathrm{~m} / \mathrm{s}\). Then the root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (A) \(836 \mathrm{~m} / \mathrm{s}\) (B) \(643 \mathrm{~m} / \mathrm{s}\) (C) \(1930 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (D) \([(1930) / \sqrt{3}] \mathrm{m} / \mathrm{s}\)

Short answer

The short answer is: The root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (B) \(643 \mathrm{~m} / \mathrm{s}\).

Step-by-step solution

Calculate the molar mass of hydrogen and Oxygen molecules.

First, we need to know the molar mass of hydrogen (H) and oxygen (O) molecules: - Hydrogen molecule (H2): 1g/mol × 2 = 2 g/mol = \(2×10^{-3}\) kg/mol - Oxygen molecule (O2): 16g/mol × 2 = 32 g/mol = \(32×10^{-3}\) kg/mol

Find the ratio of the root mean square speeds of hydrogen and oxygen molecules.

According to the root mean square speed formula: $$\frac{v_{rms_{H}}}{v_{rms_{O}}} = \sqrt{\frac{3R T_{H}}{M_{H}} \times \frac{M_{O}}{3 R T_{O}}} = \sqrt{\frac{M_O T_H}{M_H T_O}}$$ We can now substitute the given values and molar masses into the equation: $$\frac{1930}{v_{rms_{O}}} = \sqrt{\frac{(32×10^{-3})(300)}{(2×10^{-3})(900)}}$$

Calculate the root mean square speed of oxygen molecules at 900 K.

Now we need to solve for \(v_{rms_O}\): $$v_{rms_{O}}= \frac{1930}{\sqrt{\frac{(32×10^{-3})(300)}{(2×10^{-3})(900)}}}$$ Calculate the value: $$v_{rms_{O}}= 643 \mathrm{~m} / \mathrm{s}$$ So the correct answer is (B) \(643 \mathrm{~m} / \mathrm{s}\).

Key concepts

Molar Mass

The concept of molar mass is crucial when discussing the behavior of gases. Molar mass, which is the mass of a given substance divided by the amount of substance, is usually expressed in grams per mole (g/mol). For hydrogen (H₂), each atom has a molar mass of 1 g/mol, so two atoms make 2 g/mol. Convert this to kilograms per mole, resulting in \(2 \times 10^{-3}\) kg/mol.

For oxygen (O₂), each atom has a molar mass of 16 g/mol, giving O₂ a molar mass of 32 g/mol or \(32 \times 10^{-3}\) kg/mol. Molar mass is a key factor in determining how fast different gas molecules move under the same conditions.

Kinetic Theory of Gases

The kinetic theory of gases offers a simple explanation of the behavior of gases, based on the idea that gas molecules are in constant random motion. This motion is directly tied to the temperature and pressure of the gas.

• Gas molecules are considered small spheres in random motion.
• There are no significant attractive or repulsive forces between the molecules.
• The average kinetic energy of gas molecules depends solely on the temperature.

This theory explains why lighter molecules like hydrogen move faster than heavier molecules such as oxygen at the same temperature. The root mean square speed formula, \(v_{rms} = \sqrt{\frac{3RT}{M}}\), is derived from these principles.

Temperature Dependence

Temperature plays a vital role in the behavior and speed of gas molecules. According to the kinetic theory, as temperature increases, the average kinetic energy and speed of molecules also increase.

For instance, if the temperature triples, as it does from 300 K to 900 K in our problem, the speed doesn’t triple directly. Instead, it influences the root mean square speed as \(v_{rms} \propto \sqrt{T}\). This means the speed of oxygen molecules at 900 K is determined by the square root of the temperature ratio compared to 300 K, alongside changes in molar mass.

Speed of Molecules

The speed of gas molecules can be quantified in different ways, but the most common measure is the root mean square speed. This is a type of average speed that considers both mass and temperature, making it central to understanding gas behavior.

• Lighter molecules move faster than heavier ones at the same temperature.
• Changes in temperature affect the speed; increasing temperature raises it.

For example, knowing the root mean square speed of hydrogen at 300 K helps us find oxygen's speed at 900 K. By using the ratio of speeds, \(\frac{v_{rms_{H}}}{v_{rms_{O}}} = \sqrt{\frac{M_O T_H}{M_H T_O}}\), we find that oxygen's speed is 643 m/s at 900 K, showing the connection between these variables.