Question

The average kinetic energy per molecule of a gas at \(-23^{\circ} \mathrm{C}\) and \(75 \mathrm{~cm}\) pressure is \(5 \times 10^{-14}\) erg for \(\mathrm{H}_{2}\). The mean kinetic energy per molecule of the \(\mathrm{O}_{2}\) at \(227^{\circ} \mathrm{C}\) and \(150 \mathrm{~cm}\) pressure will be (A) \(80 \times 10^{-14} \mathrm{erg}\) (B) \(10 \times 10^{-14}\) erg (C) \(20 \times 10^{-14}\) erg (D) \(40 \times 10^{-14}\) erg

Short answer

The mean kinetic energy per molecule for O₂ at \( 227^{\circ} \mathrm{C} \) and \( 150 \mathrm{~cm} \) pressure is \( 10 \times 10^{-14} \mathrm{erg} \), which corresponds to option (B).

Step-by-step solution

Convert temperatures to Kelvin

We first need to convert the given temperatures from Celsius to Kelvin: For H₂: \(T_{1} = -23^{\circ} \mathrm{C} \ + 273.15 \mathrm{K} = 250.15 \mathrm{K} \) For O₂: \(T_{2} = 227^{\circ} \mathrm{C} \ + 273.15 \mathrm{K} = 500.15 \mathrm{K} \)

Use the mean kinetic energy formula

Now, we will use the formula for mean kinetic energy per molecule, \( \bar{K.E} = \frac{3}{2} k T \), to find the relationship between the given mean kinetic energy for H₂ and the mean kinetic energy we want to find for O₂: Let \( \bar{K.E_{1}}\) be the mean kinetic energy per molecule for H₂, and \( \bar{K.E_{2}}\) for O₂. We are given: \( \bar{K.E_{1}} = 5 \times 10^{-14} \mathrm{erg} \) We want to find: \( \bar{K.E_{2}} \) Using the temperatures from Step 1, we can write the mean kinetic energy for H₂ and O₂ as: \( \bar{K.E_{1}} = \frac{3}{2} k T_{1} \) \( \bar{K.E_{2}} = \frac{3}{2} k T_{2} \)

Set up a ratio and solve for O₂ mean kinetic energy

To find the mean kinetic energy per molecule for O₂, we can set up a ratio: \( \frac{\bar{K.E_{1}}}{\bar{K.E_{2}}} = \frac{T_{1}}{T_{2}} \) Now, we just need to plug in the given information and solve for \( \bar{K.E_{2}} \): \( \frac{5 \times 10^{-14} \mathrm{erg}}{\bar{K.E_{2}}} = \frac{250.15 \mathrm{K}}{500.15 \mathrm{K}} \) Rearranging to solve for \(\bar{K.E_{2}}\): \( \bar{K.E_{2}} = 5 \times 10^{-14} \mathrm{erg} \times \frac{500.15 \mathrm{K}}{250.15 \mathrm{K}} \) \( \bar{K.E_{2}} = 10 \times 10^{-14} \mathrm{erg} \) Therefore, the mean kinetic energy per molecule for O₂ at \( 227^{\circ} \mathrm{C} \) and \( 150 \mathrm{~cm} \) pressure is \( 10 \times 10^{-14} \mathrm{erg} \), which corresponds to option (B).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics, representing the relationship between pressure, volume, temperature, and the amount of gas. This law is typically expressed as \( PV = nRT \), where:

• \( P \) stands for the pressure of the gas.
• \( V \) symbolizes the volume it occupies.
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant.
• \( T \) represents the temperature measured in Kelvin.

The ideal gas law is pivotal when studying gases, as it interconnects several state variables. It's important to note that this law holds best for hypothetical gases, often referred to as 'ideal gases,' where interactions between molecules are negligible, and the size of the molecules is much smaller than the distance between them. It’s used widely across thermodynamics and kinetics to predict how gases will respond to changes in environment.

Temperature Conversion

In thermodynamics, temperature conversion is essential since temperature often needs to be expressed in Kelvin rather than Celsius or Fahrenheit. This conversion is critical because the Kelvin scale is absolute, where 0 Kelvin is the point where molecular movement theoretically stops.

• To convert from Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, \[ T_{K} = T_{C} + 273.15 \]So, \( -23^{\circ} \)C becomes 250.15 K.
• This conversion is necessary for formulas involving kinetic energy or gas laws, as they depend on absolute temperature.

Understanding how to convert temperatures correctly is crucial for solving problems involving chemical reactions, physical changes, and other processes that depend on temperature variations.

Molecular Kinetic Theory

The Molecular Kinetic Theory explains the behavior of gases, providing insight into their physical properties. This theory posits that gas molecules are in constant motion, colliding with each other and their container walls.

• The theory suggests that the temperature of a gas is directly related to the average kinetic energy of its molecules. In mathematical terms, \[ \bar{K.E} = \frac{3}{2} k T \] where \( k \) is Boltzmann's constant and \( T \) is the absolute temperature.
• Molecular motion becomes more vigorous with increasing temperature, elevating the kinetic energy and pressure exerted by the gas.
• Despite its simplicity, this theory effectively explains several gas behaviors, including temperature dependence and response to external pressure changes.

By understanding this theory, we gain insights not only into ideal gases but also into real-world applications such as engines and refrigerators.

Pressure Influence on Gases

Pressure plays a significant role in determining the state and behavior of gas molecules. According to the kinetic theory, when gas pressure increases, it usually suggests that molecules are being crowded into a smaller space, leading to more frequent collisions.

• An increase in pressure at constant temperature typically results in volume reduction, according to Boyle's Law, \[ P_1V_1 = P_2V_2 \].
• High-pressure environments can cause a gas to deviate from ideal behavior as intermolecular forces become more significant. These interactions can lead to changes in states such as compression into liquids.
• Understanding pressure's influence helps in designing equipment and processes where controlling gas behavior is crucial, like in chemical reactors or air conditioning systems.

Fields such as meteorology, engineering, and aerodynamics frequently explore pressure impacts to predict gas behavior under varied conditions.