Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The average velocity of an ideal gas molecule at \(27^{\circ} \mathrm{C}\) is \(0.3 \mathrm{~m} / \mathrm{sec}\). The average velocity at \(927^{\circ} \mathrm{C}\) will be: (a) \(0.6 \mathrm{~m} / \mathrm{sec}\) (b) \(0.3 \mathrm{~m} / \mathrm{sec}\) (c) \(0.9 \mathrm{~m} / \mathrm{sec}\) (d) \(3.0 \mathrm{~m} / \mathrm{sec}\)

Step-by-Step Solution

Answer

(a) 0.6 m/s

Step 1: Understand the Given Information

We are given the average velocity of an ideal gas molecule at a temperature of \(27^{\circ} \mathrm{C}\) as \(0.3 \mathrm{~m/s}\). We need to find the average velocity at \(927^{\circ} \mathrm{C}\). The concept we will use is that the average velocity \(v\) of an ideal gas molecule is proportional to the square root of its absolute temperature \(T\) (in Kelvin). This relationship can be expressed as \(v \propto \sqrt{T}\).

Step 2: Convert Temperatures to Kelvin

To use the velocity-temperature relationship, we first need to convert the Celsius temperatures to Kelvin using the conversion formula: \(T(K) = T(^{\circ}C) + 273.15\). - \(27^{\circ} \mathrm{C}\) becomes \(27 + 273.15 = 300.15 \mathrm{~K}\).- \(927^{\circ} \mathrm{C}\) becomes \(927 + 273.15 = 1200.15 \mathrm{~K}\).

Step 3: Apply the Proportionality Relationship

Using the formula \(v \propto \sqrt{T}\), we know that the ratio of velocities \(\frac{v_2}{v_1}\) is equal to the square root of the ratio of their respective temperatures \(\frac{T_2}{T_1}\). Therefore, \[\frac{v_2}{0.3} = \sqrt{\frac{1200.15}{300.15}}\]

Step 4: Calculate the Ratio

Compute the ratio \(\frac{1200.15}{300.15}\), which simplifies to \(\frac{1200}{300} = 4\). Take the square root of 4 to find the ratio of velocities:\[\sqrt{4} = 2\]

Step 5: Compute the New Velocity

Multiply the initial velocity \(v_1 = 0.3 \mathrm{~m/s}\) by the ratio found:\[v_2 = 0.3 \times 2 = 0.6 \mathrm{~m/s}\]

Step 6: Choose Correct Answer

Compare the computed average velocity \(0.6 \mathrm{~m/s}\) at \(927^{\circ} \mathrm{C}\) to the given options. The correct choice is (a) \(0.6 \mathrm{~m/s}\).

Key Concepts

Ideal Gas

In the realm of physics and chemistry, an ideal gas is a hypothetical gas that perfectly follows the ideal gas law, which is an equation of state of a hypothetical ideal gas. This law is a good approximation for the behavior of real gases under many conditions, although it has its limitations. The main assumption here is that the molecules do not attract or repel each other, essentially behaving as if they have no volume or interaction, except in perfect elastic collisions.
- **Properties of Ideal Gases**: - The gas consists of a large number of identical molecules. - The molecules are in constant random motion at high speeds. - All collisions between the molecules are perfectly elastic, so no energy is lost.- The ideal gas law can be described using the equation: \[ PV = nRT \] where: - \( P \) is pressure, - \( V \) is volume, - \( n \) is number of moles, - \( R \) is the ideal gas constant, - \( T \) is temperature in Kelvin.
This formula helps to understand how variables such as pressure, volume, and temperature interact in gases.

Average Velocity

The average velocity of gas molecules is essential in understanding the kinetic theory of gases. It provides insights into how fast particles are moving on average, contributing to properties like pressure and temperature. For ideal gases, the average velocity is considered to be proportional to the square root of the temperature.
- **Average Velocity Formula**: The average velocity \( v \) of an ideal gas at temperature \( T \) is expressed as: \[ v \propto \sqrt{T} \]- As temperature increases, molecules move faster, thereby increasing the average velocity.
In this exercise, the increase in temperature from 27°C to 927°C significantly impacts the average velocity of the gas molecules, allowing us to calculate how the velocity changes with temperature.

Temperature Conversion

Temperature conversion is a crucial step in many calculations involving gases, as many equations require absolute temperature measured in Kelvin. To convert from Celsius to Kelvin, you add 273.15 to the Celsius value. This step is vital for ensuring accurate and meaningful calculations in gas behavior studies.
- **Conversion Formula**: The conversion from Celsius to Kelvin is: \[ T(K) = T(^{\circ}C) + 273.15 \]
For this exercise:- 27°C converts to \[ 27 + 273.15 = 300.15 \, K \]- 927°C converts to \[ 927 + 273.15 = 1200.15 \, K \]
Accurately converting temperatures to Kelvin helps align the temperatures with the proportionality laws applied in ideal gas calculations.

Proportional Relationship

A proportional relationship in gas properties means one variable changes at a constant rate with another. For ideal gases, the velocity of molecules is directly proportional to the square root of the temperature. This means that if the temperature increases, the velocity increases as well.
- **Application in Calculations**: - Given that \( v \propto \sqrt{T} \), it indicates that the ratio of velocities \( \frac{v_2}{v_1} \) at different temperatures \( T_2 \) and \( T_1 \) is given by: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
This principle allows us to find the average velocity at any other temperature, so long as the initial velocity and temperatures are known.
For example, in our exercise going from 27°C to 927°C, this relationship helps us determine that the average velocity increases, lining up with the increased temperature when converted to the Kelvin scale.