Question

The average kinetic energy of hydrogen molecules at \(300 \mathrm{~K}\) is \(E\). At the same temperature the average kinetic energy of oxygen molecules will be (A) [E/(16)] (B) \(E\) (C) \(4 \mathrm{E}\) D) \([E /(4)]\)

Short answer

The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\). So, the correct option is (B) \(E\).

Step-by-step solution

Identify the given information

We are given: 1. The temperature of both hydrogen and oxygen gases: \(T = 300~K\) 2. The average kinetic energy of hydrogen molecules: \(E\) Our goal is to find the average kinetic energy of oxygen molecules at the same temperature.

Recall the equation for the average kinetic energy

For an ideal gas, the average kinetic energy (\(E_k\)) of a molecule is given by: \[E_k = \dfrac{3}{2}kT\] where \(k\) is Boltzmann's constant, and \(T\) is the temperature in Kelvin.

Use the equation to determine the average kinetic energy of the hydrogen molecules

We are given that the average kinetic energy of hydrogen molecules at \(300~K\) is \(E\). So, using the equation, we can write: \[E = \dfrac{3}{2}kT\]

Determine the average kinetic energy of the oxygen molecules

Since the temperature is the same for both the hydrogen and oxygen gas molecules (\(T = 300~K\)), the average kinetic energies of oxygen molecules, \(E_o\), must also be given by: \[E_o = \dfrac{3}{2}kT\]

Compare the average kinetic energies of hydrogen and oxygen molecules

As per step 3 and 4, we know that \(E = \dfrac{3}{2}kT\) and \(E_o=\dfrac{3}{2}kT\). Comparing both expressions, we get: \[E_o = E\] So, the average kinetic energy of oxygen molecules at the same temperature is the same as that of hydrogen molecules.

Answer

The correct option is: (B) \(E\) The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\).

Key concepts

Ideal Gas

An ideal gas is a hypothetical gas that perfectly follows the ideal gas law, meaning the molecules are point particles that interact only through elastic collisions. This means that ideal gases do not have real volume or forces acting between the gas molecules. These assumptions help simplify calculations, especially when exploring concepts like average kinetic energy.

The behavior of ideal gases is important when considering equations like the kinetic energy formula. In this context, we focus on how temperature and molecular motion relate. Since interactions are non-existent in an ideal gas, factors like molecular size or intermolecular forces do not alter the calculations. This makes prediction and calculation of kinetic energy much more straightforward.

Boltzmann's Constant

Boltzmann's constant (\(k\)) is a fundamental constant that connects macroscopic and microscopic physics realms. Its value is approximately \(1.38 \times 10^{-23} \text{J/K}\). This constant plays a crucial role in statistical mechanics, where it is used to relate the average kinetic energy of particles within a gas to temperature in Kelvin.

In equations like \(E_k = \dfrac{3}{2}kT\), Boltzmann's constant becomes vital. It aids in calculating the energy of each molecule in relation to temperature. Essentially, Boltzmann’s constant acts like a bridge, allowing us to connect the scale of individual particles with that of larger systems, indicating how thermal energy translates to microscopic motion.

Temperature in Kelvin

Temperature in Kelvin is the absolute scale for measuring thermal energy, distinct because it starts at absolute zero, the lowest limit of thermodynamic temperature. Zero Kelvin (\(0~K\)) marks the point where molecular motion theoretically ceases. The Kelvin scale provides a straightforward and direct method for using temperature in scientific equations, such as average kinetic energy calculations.

When dealing with gases, using Kelvin becomes essential. It directly links to how fast molecules in the gas move: higher temperatures mean more vigorous molecular motion. Since the energy equation \(E_k = \dfrac{3}{2}kT\) uses Kelvin, the relation between KE and temperature is linear and manifests in consistent units. Thus, using Kelvin allows scientists to accurately predict and compare kinetic energy irrespective of the gas type, assuming all other variables are constant.