Problem 160 - Master Resource Book in JEE Main Physics

Question

Assertion The root mean speed (rms) of oxygen molecules at a certain absolute temperature \(T\) is \(c\).If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed would be \(2 \mathrm{c}\). Reason \(c \propto \sqrt{\frac{T}{M}}\)

Step-by-Step Solution

Answer

The assertion is incorrect; the new rms speed is \( \sqrt{2}c \), not \( 2c \).

Step 1: Understanding Root Mean Square Speed

The root mean square speed (rms), denoted as \( c \), of gas molecules is given by the formula \( c = \sqrt{\frac{3kT}{M}} \), where \( k \) is Boltzmann's constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

Step 2: Evaluating the Initial Condition

Initially, the root mean square speed of oxygen molecules, which are diatomic, is \( c \) at temperature \( T \). Therefore, we can express this as \( c = \sqrt{\frac{3kT}{M_{O_2}}} \), where \( M_{O_2} \) is the molar mass of diatomic oxygen.

Step 3: Considering the Change in Temperature and Dissociation

When the temperature is doubled, \( T \) becomes \( 2T \). Additionally, if oxygen molecules dissociate into atomic oxygen, the molar mass changes to \( M_O \) (approximately half of \( M_{O_2} \) because oxygen is diatomic and becomes atomic).

Step 4: Calculating New Root Mean Square Speed for Atomic Oxygen

The new root mean square speed after temperature change and dissociation is \( \sqrt{\frac{3k(2T)}{M_O}} \). Given that \( M_O = \frac{1}{2} M_{O_2} \), substituting in gives:\[c_{new} = \sqrt{\frac{3k(2T)}{\frac{1}{2}M_{O_2}}} = \sqrt{\frac{6kT}{M_{O_2}}}\]

Step 5: Relating New Root Mean Square Speed to Original Speed

Since \( c = \sqrt{\frac{3kT}{M_{O_2}}} \), we have:\[c_{new} = \sqrt{2}c\] This calculation shows that the new speed is \( \sqrt{2} \) times the original speed, not 2c as initially asserted in the problem.

Key Concepts

Temperature Effect on Gases

The temperature of a gas has a direct effect on the motion of its molecules. When the temperature increases, the kinetic energy of the gas molecules also increases. This means the molecules move more rapidly. Temperature, often measured in Kelvin, is directly related to the root mean square (rms) speed of gas molecules by the formula:

• \( c = \sqrt{\frac{3kT}{M}} \)
• Where \( c \) is the rms speed, \( k \) is Boltzmann's constant, \( T \) is the absolute temperature, and \( M \) is the molar mass.

When the temperature of a gas doubles, the kinetic energy also doubles, resulting in an increase in the rms speed. However, the increase in rms speed due to doubled temperature is not linear; it depends on the square root, which results in the speed increasing by a factor of \( \sqrt{2} \). This is crucial to understand when considering changes in temperature and their effect on molecular speed.

Dissociation of Gases

Dissociation refers to the process where molecules split into smaller particles, such as atoms or ions. In the case of oxygen gas, at elevated temperatures, diatomic oxygen molecules \( O_2 \) can dissociate into atomic oxygen \( O \). This change affects the molar mass involved in calculating molecular speeds:

• Originally, diatomic oxygen has a molar mass \( M_{O_2} \).
• Upon dissociation, each molecule splits, effectively halving the molar mass to \( M_{O} \).

This substantial change in molar mass must be accounted for when calculating the new rms speed. Since the molar mass decreases upon dissociation, the rms speed increases further than just from the temperature change alone. New speeds are affected because molar mass is inversely proportional to the rms speed as seen in the formula: \( c \propto \sqrt{\frac{T}{M}} \). Thus, dissociation notably augments the change in molecular speed.

Molecular Speed Calculations

Calculating the speed of molecules within a gas involves considering both temperature and molecular makeup (molar mass). The root mean square speed (rms) is commonly used to describe the speed of molecules in a gas. It is derived from kinetic theory and expressed by:

• \( c = \sqrt{\frac{3kT}{M}} \)

When conducting molecular speed calculations:

• Determine the initial rms speed using the initial temperature and molar mass.
• Adjust the temperature as per the given change (such as doubling).
• Re-calculate using any change in molecular composition (like dissociation).

For the exercise, the initial speed was given as \( c \) at temperature \( T \), with a molar mass \( M_{O_2} \). Upon temperature doubling and dissociation into atomic oxygen, modifications are made to derive the new speed. The calculation effectively shows the new rms speed of \( \sqrt{2} \) times the original, highlighting how both temperature change and dissociation impact molecular speed.