Problem 110 - Master Resource Book in JEE Main Physics

Question

The average kinetic energy of a gas molecule at \(27^{\circ} \mathrm{C}\) is \(6.21 \times 10^{-21} \mathrm{~J} .\) Its average kinetic energy at \(127^{\circ} \mathrm{C}\) will be (a) \(12.2 \times 10^{-21} \mathrm{~J}\) (b) \(8.28 \times 10^{-21} \mathrm{~J}\) (c) \(10.35 \times 10^{-21} \mathrm{~J}\) (d) \(11.35 \times 10^{-21} \mathrm{~J}\)

Step-by-Step Solution

Answer

The average kinetic energy at \(127^{\circ}C\) is \(8.28 \times 10^{-21}\) J.

Step 1: Convert Celsius to Kelvin

First, convert the temperatures from Celsius to Kelvin. The formula is \( T(K) = T(^{\circ}C) + 273.15 \).\[T_1 = 27^{\circ}C + 273.15 = 300.15 \, K\]\[T_2 = 127^{\circ}C + 273.15 = 400.15 \, K\]

Step 2: Use the Relationship of Kinetic Energy and Temperature

The average kinetic energy \( E \) of a gas molecule is proportional to its temperature in Kelvin. So, \( \frac{E_1}{E_2} = \frac{T_1}{T_2} \). Given \( E_1 = 6.21 \times 10^{-21} \, J \), we want to find \( E_2 \).\[\frac{6.21 \times 10^{-21}}{E_2} = \frac{300.15}{400.15}\]

Step 3: Solve the Proportion

Rearrange the equation from Step 2 to solve for \( E_2 \).\[E_2 = \frac{6.21 \times 10^{-21} \, J \times 400.15}{300.15}\]\[E_2 \approx 8.28 \times 10^{-21} \, J\]

Step 4: Select the Correct Option

Compare the calculated \( E_2 \) with the given options. The closest match is:(b) \( 8.28 \times 10^{-21} \, J \).

Key Concepts

Temperature Conversion

Temperature conversion is an essential concept, especially when dealing with gas laws and kinetic energy of molecules. In scientific calculations, temperatures are often measured in Kelvin, the absolute temperature scale, rather than Celsius or Fahrenheit.
To convert a temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature:

• For 27°C, the conversion is: \[T(K) = 27 + 273.15 = 300.15 \ K\]
• For 127°C, it becomes: \[T(K) = 127 + 273.15 = 400.15 \ K\]

This conversion is crucial for using the Ideal Gas Law and when discussing kinetic energy, as these concepts rely on temperature measured in Kelvin, which is proportional directly to the energy of gas molecules.

Proportional Relationships

The concept of proportional relationships emerges frequently in physics, especially with gas molecules' kinetic energy and temperature. When two quantities are proportional, one varies directly as the other. This means when one value increases, the other one does too, by a consistent factor.
In gases, the average kinetic energy ( \(E\)) of molecules is proportional to the temperature ( \(T\)) in Kelvin: \[ \frac{E_1}{E_2} = \frac{T_1}{T_2}\]This equation allows us to predict changes in kinetic energy based on temperature changes. In our example, the initial kinetic energy at 300.15 K is given, and we solve for the new kinetic energy when the temperature rises to 400.15 K. By understanding this relationship, you can determine the energy changes that occur with temperature variations, which is valuable for interpreting a wide range of physical systems.

Ideal Gas Law

The Ideal Gas Law is a well-known equation in chemistry and physics that describes the behavior of an ideal gas. Though our problem focuses on kinetic energy, understanding this law is essential to grasping why temperatures and pressures affect gases as they do. The Ideal Gas Law is given by: \[PV = nRT\]Where:

• P is the pressure of the gas,
• V is the volume of the gas,
• n is the number of moles of the gas,
• R is the ideal gas constant, and
• T is the absolute temperature in Kelvin.

This equation can be rearranged to understand different aspects of gas behavior. In context with kinetic energy, at constant volume and number of moles, if the temperature increases, so does the pressure or kinetic energy. Hence, the Ideal Gas Law provides the framework to connect kinetic energy with temperature changes in gases, complementing the proportional relationships concept.