Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at \(300 \mathrm{~K}\) is: Atomic masses: He \(=4 \mathrm{u}, \mathrm{O}=16 \mathrm{u})\) (a) \(300 \mathrm{~K}\) (b) \(600 \mathrm{~K}\) (c) \(1200 \mathrm{~K}\) (d) \(2400 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is 2400 K, option (d).

Step 1: Understand Root Mean Square Speed

The root mean square (RMS) speed 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 mass of the gas particle.

Step 2: Set Up the Equation for RMS Speeds

Since the RMS speeds of Helium and Oxygen should be the same, we equate their RMS speed equations: \( \sqrt{\frac{3kT_{ ext{O}_2}}{m_{ ext{O}_2}}} = \sqrt{\frac{3kT_{ ext{He}}}{m_{ ext{He}}}} \). Here, \( T_{ ext{O}_2} \) is the temperature for oxygen, and \( T_{ ext{He}} = 300 \text{ K} \) for helium.

Step 3: Simplify and Solve for the Oxygen Temperature

Squaring both sides removes the square root: \( \frac{3kT_{ ext{O}_2}}{m_{ ext{O}_2}} = \frac{3kT_{ ext{He}}}{m_{ ext{He}}} \). Canceling \(3k\) from both sides gives \( \frac{T_{ ext{O}_2}}{m_{ ext{O}_2}} = \frac{T_{ ext{He}}}{m_{ ext{He}}} \).

Step 4: Substitute Atomic Masses and Known Temperature

The molar mass of an oxygen molecule \( \text{O}_2 \) is \( 2 \times 16 = 32 \text{ u} \). Substitute \( m_{ ext{O}_2} = 32 \) u, \( m_{ ext{He}} = 4 \) u, and \( T_{ ext{He}} = 300 \text{ K} \) into the equation: \( \frac{T_{ ext{O}_2}}{32} = \frac{300}{4} \).

Step 5: Solve for Temperature of Oxygen (\( T_{\text{O}_2} \))

Cross-multiply to solve for \( T_{\text{O}_2} \): \( T_{ ext{O}_2} = 32 \times \frac{300}{4} \). Calculating this gives \( T_{ ext{O}_2} = 2400 \text{ K} \).

Key Concepts

Root Mean Square Speed

In the world of gases, the root mean square (RMS) speed is a fundamental concept. It provides us with a mathematical means to describe the speed of molecules within a sample of gas, taking into account their randomness and different speeds.
The RMS speed is determined by the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]where:

• \(v_{rms}\) is the root mean square speed,
• \(k\) is the Boltzmann constant, a proportionality factor,
• \(T\) is the absolute temperature (measured in Kelvin), and
• \(m\) is the mass of a single gas particle.

Temperature and mass both play crucial roles. As temperature rises, molecules gain energy and move faster, hence increasing the RMS speed. Conversely, as mass increases, the RMS speed generally decreases because it requires more energy to accelerate heavier particles. Understanding RMS speed is vital as it links particle kinetics to macroscopic properties like pressure and temperature.

Atomic Mass Calculation

Atomic mass is essential for understanding the behavior of gases. It's a measure of the mass of an atom, typically expressed in unified atomic mass units (u). For molecules, which are combinations of atoms, the molecular mass is the sum of the individual atomic masses contributing to that molecule.
For instance, in this exercise, we consider helium (He) and oxygen (O\(_2\)). Helium, being a noble gas, consists of single atoms with an atomic mass of 4 u per atom. Oxygen, a diatomic molecule in natural conditions, consists of two oxygen atoms each with an atomic mass of 16 u, hence its molecular mass is calculated as:\[ m_{\text{O}_2} = 16 \text{ u} \times 2 = 32 \text{ u} \]Accurate mass calculations are crucial as they directly influence the RMS speed and other kinetic properties of gases. Knowing the atomic or molecular mass allows us to predict and calculate how gases will behave under various conditions.

Temperature and Molecular Speed

Temperature is intricately linked to how fast molecules move. Higher temperatures mean more energy is available, so molecules move faster. When studying gases, this concept is critical because it affects things like reaction rates and pressure.
The relationship between temperature and molecular speed is highlighted through the equation for RMS speed, showing that speed is directly dependent on the temperature of a gas. For gases at the same RMS speed but different temperatures, such as helium at 300 K and oxygen, we adjust the temperature to compensate for differences in mass.
By equating RMS speeds and calculating the necessary temperature for oxygen to match helium's speed:\[ \frac{T_{\text{O}_2}}{m_{\text{O}_2}} = \frac{T_{\text{He}}}{m_{\text{He}}} \]where \(T_{\text{He}}\) is 300 K for helium and \(m_{\text{He}} = 4 \text{ u}\), you arrive at the solution for oxygen. By multiplying the temperature ratio by the oxygen molecular mass, we find the required temperature for oxygen molecules to achieve the same RMS speed as helium:\[ T_{\text{O}_2} = 2400 \text{ K} \]This showcases how closely tied molecular speed is to temperature and reinforces the need to understand these connections fully.