Problem 41 - Chemistry and Chemical Reactivity

Question

You have two flasks of equal volume. Flask A contains \(\mathrm{H}_{2}\) at \(0^{\circ} \mathrm{C}\) and 1 atm pressure. Flask \(\mathrm{B}\) contains \(\mathrm{CO}_{2}\) gas at \(25^{\circ} \mathrm{C}\) and 2 atm pressure. Compare these two gases with respect to each of the following: (a) average kinetic energy per molecule (b) root mean square speed (c) number of molecules (d) mass of gas

Step-by-Step Solution

Answer

(a) Flask B has higher average kinetic energy. (b) Flask A has higher root mean square speed. (c) Flask B has more molecules. (d) Flask B has more mass of gas.

Step 1: Understanding Average Kinetic Energy

The average kinetic energy per molecule of a gas depends solely on the temperature in Kelvin, not on the type of gas or its pressure. The formula for average kinetic energy per molecule is \( KE = \frac{3}{2} kT \), where \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin. For Flask A, \( T = 273.15 \) K (from \( 0^{\circ} \mathrm{C}\)), and for Flask B, \( T = 298.15 \) K (from \( 25^{\circ} \mathrm{C} \)). Thus, Flask B will have a higher average kinetic energy per molecule than Flask A.

Step 2: Calculating Root Mean Square Speed

The root mean square speed \( (v_{rms}) \) is calculated using \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the ideal gas constant, \( T \) is in Kelvin, and \( M \) is the molar mass in kilograms/mol. For \( \mathrm{H}_2 \) in Flask A, \( M = 2.016 \times 10^{-3} \) kg/mol. For \( \mathrm{CO}_2 \) in Flask B, \( M = 44.01 \times 10^{-3} \) kg/mol. Although Flask B is at a higher temperature, \( \mathrm{H}_2 \) has a much lower molar mass, leading to a higher \( v_{rms} \) for \( \mathrm{H}_2 \).

Step 3: Counting Number of Molecules

To determine the number of molecules, utilize the ideal gas law \( PV = nRT \), where \( n \) is the number of moles. Given constant volume, we can set the moles \( n \) proportional to \( \frac{P}{T} \). Flask A's ratio is \( \frac{1}{273.15} \) and Flask B's ratio is \( \frac{2}{298.15} \). Thus, Flask B contains more gas molecules than Flask A due to its higher \( \frac{P}{T} \) ratio.

Step 4: Determining Mass of Gas

The mass of gas depends on the number of moles and molar mass (\( m = nM \)). Given that Flask B has more molecules, it also contains more moles of gas. Since \( \mathrm{CO}_2 \) has a higher molar mass than \( \mathrm{H}_2 \), Flask B has a greater mass of gas.

Key Concepts

Kinetic Energy

Kinetic energy is a measure of how fast molecules are moving. For gases, average kinetic energy per molecule is determined only by temperature, expressed in Kelvin. The formula is \( KE = \frac{3}{2} kT \), where \( k \) is the Boltzmann constant, and \( T \) is the temperature. The type or pressure of the gas does not affect this energy. For the given exercise, Flask A has a temperature of 273.15 K while Flask B is at 298.15 K. This means Flask B has a higher average kinetic energy per molecule because it is at a higher temperature. Keep in mind, more kinetic energy means more movement at the molecular level.

Root Mean Square Speed

Root mean square speed \( (v_{rms}) \) is a way to express the average speed of gas molecules. It's calculated with the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the ideal gas constant, \( T \) is temperature, and \( M \) is the molar mass. Even if the temperature is higher in Flask B, the molar mass of \( \text{H}_2 \) in Flask A is much lower than \( \text{CO}_2 \) in Flask B. This results in a higher \( v_{rms} \) for \( \text{H}_2 \) because lighter molecules move faster than heavier ones at the same energy level. This reflects the balance between mass and speed in gas behavior.

Ideal Gas Law

The ideal gas law \( PV = nRT \) is fundamental in understanding gases. It connects pressure \( P \), volume \( V \), the number of moles \( n \), the gas constant \( R \), and temperature \( T \). To find the number of molecules, one can use the relation \( \frac{P}{T} \) as proportional to moles \( n \). In the exercise, Flask A at \( P = 1 \) atm and \( T = 273.15 \) K has a ratio of \( \frac{1}{273.15} \), whereas Flask B at \( P = 2 \) atm and \( T = 298.15 \) K has \( \frac{2}{298.15} \). Thus, Flask B has a greater quantity of molecules, owing to a larger \( \frac{P}{T} \) ratio. This shows how pressure and temperature affect the number of gas molecules.

Molar Mass

Molar mass refers to the mass of one mole of a substance. It affects various properties of gases, including their mass and speed. In the exercise context, \( \text{H}_2 \) has a low molar mass of 2.016 g/mol compared to \( \text{CO}_2 \) at 44.01 g/mol. Although Flask B has more gas molecules, combining this with its higher molar mass results in a greater total mass of gas in Flask B compared to Flask A. Molar mass influences how quickly gas molecules move and how much space they occupy, impacting both physical and chemical behavior in reactions.