Animated Solution for Physics - Thermodynamics: The average translational energy and the rms speed of molecules in a sample of oxygen gas at $300$ K are $6.21\times 10^{-21}$ J and $484$ m/s respectively. The corresponding values at $600$ K are nearly (assuming ideal gas behaviour)

The behavior of gases at the microscopic level is a fascinating dance of molecules. When we talk about the temperature of a gas, we are essentially talking about how energetically these molecules are moving. In this problem, we explore how the average translational kinetic energy and the root mean square (rms) speed of oxygen molecules change when the temperature is doubled.

We are given a sample of oxygen gas at an initial temperature of $T_1=300\text{ K}$. At this state, the molecules have an average translational kinetic energy $E_1=6.21\times 10^{-21}\text{ J}$ and an rms speed $v_1=484\text{ m/s}$.

The gas is then heated to a final temperature of $T_2=600\text{ K}$. Our goal is to find the new average translational kinetic energy $E_2$ and the new rms speed $v_2$. The key to solving this lies in understanding the mathematical relationships between these microscopic properties and the macroscopic temperature.

Let's first focus on the average translational kinetic energy. According to the kinetic theory of gases, the average translational kinetic energy $E$ of a single gas molecule is given by the equation:

where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.

This equation is profoundly beautiful because it tells us that the average translational kinetic energy depends only on the absolute temperature. It doesn't matter if the gas is oxygen, hydrogen, or carbon dioxide; at a given temperature, their molecules will have the exact same average translational kinetic energy.

From this equation, we can clearly see the direct proportionality:

Since the kinetic energy is directly proportional to the temperature, any fractional change in temperature will result in the exact same fractional change in kinetic energy.

In our problem, the temperature is increased from $300\text{ K}$ to $600\text{ K}$. This means the temperature has exactly doubled:

Because $E\propto T$, the kinetic energy must also double:

Now, we simply substitute the given initial value:

This gives us the first part of our answer.

Next, let's determine the new root mean square (rms) speed. The rms speed $v_{rms}$ of gas molecules is given by the formula:

where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.

Unlike kinetic energy, the rms speed does depend on the mass of the gas molecules. However, since we are dealing with the same sample of oxygen gas, the molar mass $M$ remains constant. Therefore, the relationship between the rms speed and temperature is:

This proportionality tells us that the rms speed does not scale linearly with temperature. Instead, it scales with the square root of the temperature.

Since the temperature has doubled ($T_2=2T_1$), the new rms speed will be:

We are given the initial speed $v_1=484\text{ m/s}$. The square root of $2$ is approximately $1.414$. Substituting these values:

By understanding the fundamental proportionalities derived from the kinetic theory of gases, we were able to quickly determine the new state of the gas. The kinetic energy doubled because it is directly proportional to temperature, while the rms speed increased by a factor of $\sqrt{2}$ because it is proportional to the square root of temperature.

Our calculated values are $E_2=12.42\times 10^{-21}\text{ J}$ and $v_2=684\text{ m/s}$. Comparing this with the given options, we find that option (d) is the correct answer.