A mixture of ethane and propanol gases is maintained at a temperature of 400 K. Which of the followi — States of Matter Chemistry Question

STEPS:

1. Relate temperature to average kinetic energy:
According to the Kinetic Molecular Theory of gases, the average translational kinetic energy ($KE_{\text{avg}}$) of a gas sample is directly proportional to its absolute temperature in Kelvin ($KE_{\text{avg}} \propto T$). Since both ethane and propanol are in a gas mixture maintained at the same temperature of $400\text{ K}$, the average kinetic energy of the ethane molecules is exactly equal to that of the propanol molecules.

2. Connect kinetic energy, mass, and velocity:
The kinetic energy of an individual moving particle is given by the formula:
$$KE = \frac{1}{2}mv^2$$
where $m$ is the mass of the particle and $v$ is its speed. Since the average kinetic energies of both gas species are equal, their masses and average molecular speeds are inversely related:
$$\frac{1}{2}m_{\text{ethane}}(v_{\text{ethane}})^2 = \frac{1}{2}m_{\text{propanol}}(v_{\text{propanol}})^2$$

3. Compare the molar masses of the two gases:
From the exam specifications and the provided combustion table:
* Ethane ($\text{C}_2\text{H}_6$): Molar mass is $30.\text{ g/mol}$.
* Propanol ($\text{C}_3\text{H}_8\text{O}$): Molar mass is $60.\text{ g/mol}$.
* Ethane molecules are significantly less massive than propanol molecules.

4. Determine the relative average molecular speeds:
Because ethane molecules are lighter (less massive) than propanol molecules, they must move at a greater average molecular speed to maintain the same average kinetic energy as the heavier propanol molecules at $400\text{ K}$. This concludes that the average kinetic energies are equal, and the average speed of ethane is greater than that of propanol, making Option B the correct answer.

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WHY_OTHERS_WRONG:

• Option A is wrong because it states that the average speed of ethane is *less* than that of propanol. Because ethane molecules are lighter, they must have a higher average speed to have the same kinetic energy as the heavier propanol molecules.
• Option C is wrong because it states that the average kinetic energy of ethane is *greater* than that of propanol. Because both gases are at the exact same temperature ($400\text{ K}$), their average kinetic energies must be equal.
• Option D is wrong because the total pressure of the gas mixture is irrelevant to comparing average molecular speeds. Average speed is determined solely by temperature and molecular mass, both of which are known.