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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

Short Answer

Expert verified
(a) CO2 has higher kinetic energy; (b) H2 has greater speed; (c) H2 has more molecules; (d) CO2 is heavier.

Step by step solution

01

Understand Average Kinetic Energy

The average kinetic energy of a gas molecule is given by the formula \( KE = \frac{3}{2} k T \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. Since the temperature is the only variable in the equation, we'll determine how the temperature affects each gas. Convert the temperatures: \( 0^{\circ}C = 273 \, K \) and \( 25^{\circ}C = 298 \, K \). Since \( CO_2 \) (in Flask B) is at a higher temperature than \( H_2 \) (in Flask A), the average kinetic energy of \( CO_2 \) molecules is greater.
02

Calculate Root Mean Square Speed

The root mean square speed \( u_{rms} \) of a gas is given by \( u_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. For \( H_2 \), \( M = 2 \times 10^{-3} \) kg/mol, and \( T = 273 \, K \). For \( CO_2 \), \( M = 44 \times 10^{-3} \) kg/mol, and \( T = 298 \, K \). Calculate \( u_{rms} \) for each gas. The lighter the gas and the higher the temperature, the greater the \( u_{rms} \). \( H_2 \) has a much higher \( u_{rms} \) because its molar mass is smaller.
03

Determine the Number of Molecules

Use the ideal gas law \( PV = nRT \) to find the number of moles, \( n \), for each gas. Since both flasks have the same volume and use the same \( R \), the number of moles is determined by \( \frac{P}{T} \). Calculate: \( n_{H_2} = \frac{1}{273} \) and \( n_{CO_2} = \frac{2}{298} \). Convert these to the number of molecules using Avogadro's number. \( H_2 \) has more molecules because \( \frac{1}{273} > \frac{2}{298} \).
04

Assess the Mass of Gas

To find the mass, use \( m = nM \), where \( n \) is the number of moles and \( M \) is the molar mass. Since \( CO_2 \) has a higher pressure and molar mass, calculate \( m_{H_2} = n_{H_2} \cdot 2 \times 10^{-3} \) kg/mol and \( m_{CO_2} = n_{CO_2} \cdot 44 \times 10^{-3} \) kg/mol. The mass of \( CO_2 \) is greater than that of \( H_2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Kinetic Energy
The average kinetic energy of a gas molecule is a pivotal concept in understanding gas behavior. For any gas, this energy can be calculated using the formula \( KE = \frac{3}{2} k T \), where \( k \) stands for the Boltzmann constant, and \( T \) represents the temperature in Kelvin. Because the kinetic energy of the molecules depends directly on temperature, it is independent of the type of gas.

In practical terms, higher temperature means higher average kinetic energy for the gas molecules. Let's apply this idea to our gases. For Flask A containing \( \mathrm{H}_2 \) at \( 0^{\circ} \mathrm{C}\) (converted to 273 K), and Flask B containing \( \mathrm{CO}_2 \) at \( 25^{\circ} \mathrm{C}\) (which is 298 K), it's easy to see that the \( \mathrm{CO}_2 \) molecules have higher average kinetic energy because of the higher temperature.

Remember, the type of gas doesn’t matter here—only the temperature does.
  • Average kinetic energy is directly proportional to temperature.
  • Type of gas doesn't affect average kinetic energy.
Root Mean Square Speed
Root mean square speed (RMS speed) helps us grasp how fast gas molecules are moving. The formula used is \( u_{rms} = \sqrt{\frac{3RT}{M}} \), with \( R \) as the universal gas constant, \( T \) as the temperature in Kelvin, and \( M \) as the molar mass of the gas.

This indicates that the RMS speed is dependent on both temperature and the molar mass of the gas. Gases with lower molar masses have higher RMS speeds, given the same temperature. With our example gases, \( \mathrm{H}_2 \) with a much lighter molar mass (2 g/mol) compared to \( \mathrm{CO}_2 \) (44 g/mol), it achieves a significantly higher RMS speed, even though it is at a lower temperature (273 K compared to 298 K for \( \mathrm{CO}_2 \)).

  • RMS speed increases with decreasing molar mass.
  • Higher temperatures boost RMS speed.
Thus, even though \( \mathrm{CO}_2 \) is at a higher temperature, \( \mathrm{H}_2 \)'s significantly lighter mass means its molecules are moving much faster.
Ideal Gas Law
The ideal gas law \( PV = nRT \) serves as a fundamental equation in describing gas behaviors. It relates the pressure (\( P \)), volume (\( V \)), number of moles (\( n \)), universal gas constant (\( R \)), and temperature (\( T \)). With our scenario of two equal volume flasks, we use this law to find the number of moles of gas in each flask. Since volume is constant, the number of moles is basically determined by \( \frac{P}{T} \), keeping \( R \) consistent for both gases.

For \( \mathrm{H}_2 \), the calculation is \( n_{H_2} = \frac{1}{273} \), and for \( \mathrm{CO}_2 \), it's \( n_{CO_2} = \frac{2}{298} \). When comparing these values, \( \mathrm{H}_2 \) has more molecules due to its higher \( \frac{P}{T} \) ratio, as derived from the ideal gas law.

  • The ideal gas law helps to determine the amount of gas present under specified conditions.
  • For equal volumes, the pressure to temperature ratio determines the number of moles.
By understanding the ideal gas law, students can appreciate how gas laws provide insights into molecular quantity and behavior under various conditions.
Molar Mass
Molar mass, the mass of one mole of a substance, is crucial in calculating the extent and behavior of gases. Its typical units are grams per mole (g/mol), and it ties directly to both the number of moles and total mass of the gas.

In this example, \( \mathrm{H}_2 \) has a molar mass of 2 g/mol, whereas \( \mathrm{CO}_2 \) has a molar mass of 44 g/mol. Given a certain number of molecules (or moles), you can determine the mass of the gas using the relationship \( m = nM \), where \( n \) is the number of moles and \( M \) is the molar mass.
  • Lighter molar mass allows for faster molecular speeds but less overall gas mass.
  • Heavier molar mass means slower molecular speeds for a given temperature but greater gas mass for any given number of molecules.
The greater the molar mass, the more massive the gas will be if both \( n \) and \( M \) are determinant. Hence, although \( \mathrm{H}_2 \) presents a larger number of molecules, \( \mathrm{CO}_2 \) results in a greater overall mass due to its higher molar mass.

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Most popular questions from this chapter

You are given 1.56 g of a mixture of \(\mathrm{KClO}_{3}\) and KCl. When heated, the \(\mathrm{KClO}_{3}\) decomposes to KCl and \(\mathbf{O}_{2}\) $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ and \(327 \mathrm{mL}\) of \(\mathrm{O}_{2}\) with a pressure of \(735 \mathrm{mm} \mathrm{Hg}\) is collected at \(19^{\circ} \mathrm{C}\). What is the weight percentage of \(\mathrm{KClO}_{3}\) in the sample?

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A flask is first evacuated so that it contains no gas at all. Then, \(2.2 \mathrm{g}\) of \(\mathrm{CO}_{2}\) is introduced into the flask. On warming to \(22^{\circ} \mathrm{C},\) the gas exerts a pressure of \(318 \mathrm{mm}\) Hg. What is the volume of the flask?

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5\. A self-contained underwater breathing apparatus (SCUBA) uses canisters containing potassium superoxide. The superoxide consumes the \(\mathrm{CO}_{2}\) exhaled by a person and replaces it with oxygen. $$ 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ What mass of \(\mathrm{KO}_{2}\), in grams, is required to react with \(8.90 \mathrm{L}\) of \(\mathrm{CO}_{2}\) at \(22.0^{\circ} \mathrm{C}\) and \(767 \mathrm{mm} \mathrm{Hg} ?\)

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