Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

The most probable speeds of the molecules of gas \(\mathrm{A}\) at \(\mathrm{T}_{1} \mathrm{~K}\) and gas \(\mathrm{B}\) at \(\mathrm{T}_{2} \mathrm{~K}\) are in the ratio \(0.715: 1\). The same ratio for gas \(\mathrm{A}\) at \(\mathrm{T}_{2} \mathrm{~K}\) and gas \(\mathrm{B} \mathrm{T}_{1} \mathrm{~K}\) is \(0.954\). Find the ratio of molar masses \(\mathrm{M}_{\mathrm{A}}: \mathrm{M}_{\mathrm{B}}\). (a) \(1.965\) (b) \(1.0666\) (c) \(1.987\) (d) \(1.466\)

Step-by-Step Solution

Answer

The ratio of molar masses \( M_A:M_B \) is 1.965.

Step 1: Understanding the Formula for Most Probable Speed

The most probable speed of gas molecules is given by the formula \( v_p = \sqrt{\dfrac{2kT}{m}} \), where \( v_p \) is the most probable speed, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of a molecule. For molar mass \( M \), \( v_p = \sqrt{\dfrac{2RT}{M}} \), where \( R \) is the gas constant and \( M \) is molar mass.

Step 2: Setting Up the Equations for Both Conditions

From the problem, we have two ratios:\( \dfrac{v_{pA}(T_1)}{v_{pB}(T_2)} = 0.715 \) and \( \dfrac{v_{pA}(T_2)}{v_{pB}(T_1)} = 0.954 \). Using \( v_p = \sqrt{\dfrac{2RT}{M}} \), the first condition gives us \( \sqrt{\dfrac{T_1}{M_A} \cdot \dfrac{M_B}{T_2}} = 0.715 \), and the second condition gives \( \sqrt{\dfrac{T_2}{M_A} \cdot \dfrac{M_B}{T_1}} = 0.954 \).

Step 3: Squaring Both Equations to Remove the Square Root

Squaring both sides of the equations removes the square root, yielding: \( \dfrac{T_1 M_B}{T_2 M_A} = 0.715^2 \) and \( \dfrac{T_2 M_B}{T_1 M_A} = 0.954^2 \).

Step 4: Solving for the Ratio of Molar Masses

By dividing one equation by the other, we can find the ratio \( \dfrac{T_1^2}{T_2^2} = \dfrac{0.715^2}{0.954^2} \). This simplifies to \( \dfrac{T_1}{T_2} = \dfrac{0.715}{0.954} \), and then using \( \dfrac{M_A}{M_B} = \left(\dfrac{0.954^2 \cdot 0.715}{0.715^2 \cdot 0.954}\right) \), we compute this value to find \( M_A:M_B \).

Step 5: Calculating the Ratio of Molar Masses

Substitute the numerical values into the derived expression: \( M_A:M_B = \left(\dfrac{(0.954)^2}{(0.715)^2}\right) = 1.965 \).

Key Concepts

Most Probable Speed

In the kinetic theory of gases, the most probable speed refers to the speed at which the maximum number of gas molecules are moving. It's important to note that this speed is different from average or root-mean-square speeds.The formula to calculate the most probable speed is given by:\[ v_p = \sqrt{\frac{2kT}{m}} \]Here:

• \( v_p \) is the most probable speed,
• \( k \) represents the Boltzmann constant,
• \( T \) is the absolute temperature,
• \( m \) is the mass of a single molecule.

When working with molar mass \( M \) instead of the molecular mass, the formula adjusts to:\[ v_p = \sqrt{\frac{2RT}{M}} \]In this version:

• \( R \) is the universal gas constant, used because we are considering a mole of molecules instead of individual molecules,
• \( M \) refers to the molar mass of the gas.

The most probable speed is crucial when examining the behavior of gas molecules under different conditions of temperature and pressure, as it helps to predict how gases will react at a molecular level.

Boltzmann Constant

The Boltzmann constant is a fundamental physical constant that plays a vital role in the field of thermodynamics and statistical mechanics.It links the average kinetic energy of particles in a gas with the temperature of that gas. The value of the Boltzmann constant \( k \) is approximately \( 1.38 \times 10^{-23} \) J/K.The significance of the Boltzmann constant can be summarized as follows:

• It helps in understanding the relationship between microscopic and macroscopic phenomena. In other words, it connects properties at the atomic level to observable bulk properties.
• In the context of gases, it is a crucial element in calculating speeds, pressures, and other dynamics via equations such as the ideal gas law and kinetic energy formulas.

The constant is central in deriving formulas, such as the most probable speed, which showcases its application in determining gas molecules' velocity based on temperature.

Molar Mass Ratio

The molar mass ratio allows us to express how the mass of one mole of a substance compares to another. This ratio is pertinent, especially in gas behavior studies where different gases are compared under various conditions.By comparing the molar masses of gases, we can determine their relative behavior concerning molecular speeds.When we look at the exercise solution, we understand that by setting up equations based on the given problem and solving them, the molar mass ratio is extracted using such comparisons in molecular speed:\[ \frac{M_A}{M_B} = \left( \frac{T_1M_B}{T_2M_A} \right) \]This equation helps tie the physical properties of gases back to their molecular characteristics. Ratio comparisons like these allow students and chemists alike to appreciate the interplay between different variables in thermodynamics. Understanding this concept is crucial as it also determines how gases of different masses diffuse and interact, helping predict their behavior under varying thermal conditions.