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Calculate the rms speed for CO molecules at \(25^{\circ} \mathrm{C}\) What is the ratio of this speed to that of Ar atoms at the same temperature?

Short Answer

Expert verified
The ratio of the rms speed of CO to Ar at \(25^{\circ}\)C is approximately \(\sqrt{39.95/28.01}\).

Step by step solution

01

Understand RMS Speed Formula

The root-mean-square (rms) speed of a gas is calculated using the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where \( k \) is Boltzmann's constant \((1.38 \times 10^{-23} \text{ J/K})\), \( T \) is the temperature in Kelvin, and \( m \) is the mass of a molecule in kilograms.
02

Convert Temperature to Kelvin

Convert the given temperature into Kelvin by adding 273.15 to the Celsius temperature:\[ T = 25^{\circ} C + 273.15 = 298.15 \, K \]
03

Calculate the Molar Mass of CO and Ar

The molar mass of CO is the sum of the molar masses of carbon and oxygen: \[ \text{Molar mass of CO} = 12.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 28.01 \, \text{g/mol} \]For Ar (argon), it is approximately 39.95 g/mol.
04

Convert Molar Mass to Kilograms per Molecule

To convert molar mass to kilograms per molecule, divide by Avogadro's number \((6.022 \times 10^{23} \, \text{molecules/mol})\): For CO, \[ m_{CO} = \frac{28.01 \, \text{g/mol}}{6.022 \times 10^{23}} = \frac{28.01 \times 10^{-3}}{6.022 \times 10^{23}} \, \text{kg} \]For Ar, \[ m_{Ar} = \frac{39.95 \, \text{g/mol}}{6.022 \times 10^{23}} = \frac{39.95 \times 10^{-3}}{6.022 \times 10^{23}} \, \text{kg} \]
05

Calculate RMS Speed for CO

Insert the values from previous steps into the rms speed formula for CO:\[ v_{rms, CO} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 298.15}{\frac{28.01 \times 10^{-3}}{6.022 \times 10^{23}}}} \]
06

Calculate RMS Speed for Ar

Insert the values from previous steps into the rms speed formula for Ar:\[ v_{rms, Ar} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 298.15}{\frac{39.95 \times 10^{-3}}{6.022 \times 10^{23}}}} \]
07

Compute the Ratio of Speeds

Calculate the ratio of the rms speed of CO molecules to that of Ar atoms:\[ \text{Ratio} = \frac{v_{rms, CO}}{v_{rms, Ar}} \]

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Key Concepts

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

Kinetic Molecular Theory
The kinetic molecular theory is an essential concept for understanding the behavior of gases. It explains how the microscopic properties of gas molecules result in observable properties such as pressure and temperature. According to this theory:
  • Gases are composed of a large number of particles that are in constant random motion.
  • The particles are small and occupy a negligible volume compared to the volume of their container.
  • The collisions between gas particles, and between particles and the walls of the container, are perfectly elastic. This means that there is no loss of kinetic energy during the collisions.
  • The average kinetic energy of the gas particles is directly proportional to the temperature of the gas in Kelvin. This explains the connection between kinetic energy, temperature, and gas motion.
To calculate the root-mean-square (rms) speed of gas particles, we rely on these principles. The rms speed gives us insight into the speed distribution of the molecules in a gas sample at a specific temperature.
Gas Laws
Gas laws describe the relationships between the pressure, volume, and temperature of a gas. These laws help us understand how gases behave under varying conditions. The most commonly used gas laws are:
  • Boyle's Law: This states that the pressure of a gas is inversely proportional to its volume at a constant temperature.
  • Charles's Law: This law states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure remains constant.
  • Avogadro's Law: It indicates that the volume of a gas is directly proportional to the number of moles at constant temperature and pressure.
  • Ideal Gas Law: Combining the above laws, this law is expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
An understanding of these laws enables us to predict how gases will respond to changes in temperature, volume, and pressure. For calculating rms speed, knowing the temperature and knowing how it affects molecular motion is crucial.
Molar Mass
Molar mass is a fundamental concept in chemistry that relates the mass of a chemical substance to the amount of substance in moles. It is typically expressed in units of grams per mole (g/mol). Understanding molar mass is crucial for:
  • Converting between grams and moles, a necessary step in stoichiometry calculations.
  • Calculating the mass of a single molecule, which can be derived by dividing by Avogadro's number \(6.022 \times 10^{23}\) to convert molar mass to mass in kilograms per molecule.
  • Using the rms speed formula, where the mass of a gas particle in kilograms is needed.
For the exercise, the molar masses of CO and Ar are calculated to find their respective molecular masses, which are then used in the rms speed formula. This allows us to compare the speeds of these gas particles at the same temperature.
Temperature Conversion
Temperature conversion is a simple yet important task in chemistry, especially when working with gases. Most gas-related calculations require temperature to be expressed in Kelvin.
  • The Celsius scale, although familiar, doesn't align with the absolute temperature that is needed for kinetic molecular discussions. This is why we convert to Kelvin.
  • The conversion is straightforward: add 273.15 to the Celsius temperature to get Kelvin.
For example, to convert \(25^{\circ}C\) to Kelvin, you calculate \(25 + 273.15 = 298.15 \, K\). This ensures that your temperature accurately represents the energy of motion in a gas sample. Using Kelvin allows for consistent results when utilizing gas laws and calculating rms speeds.

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Most popular questions from this chapter

A You have a sample of helium gas at \(-33^{\circ} \mathrm{C},\) and you want to increase the rms speed of helium atoms by \(10.0 \% .\) To what temperature should the gas be heated to accomplish this?

The ideal gas law is least accurate under conditions of high pressure and low temperature. In those situations, using the van der Waals equation is advisable. (a) Calculate the pressure exerted by \(12.0 \mathrm{g}\) of \(\mathrm{CO}_{2}\) in a \(500-\mathrm{mL}\) vessel at \(298 \mathrm{K},\) using the ideal gas equation. Then, recalculate the pressure using the van der Waals equation. Assuming the pressure calculated from van der Waal's equation is correct, what is the percent error in the answer when using the ideal gas equation? (b) Next, cool this sample to \(-70^{\circ} \mathrm{C}\). Then perform the same calculation for the pressure exerted by \(\mathrm{CO}_{2}\) at this new temperature, using both the ideal gas law and the van der Waals equation. Again, what is the percent error when using the ideal gas equation?

A sample of nitrogen gas has a pressure of \(67.5 \mathrm{mm} \mathrm{Hg}\) in a 500 -mL. flask. What is the pressure of this gas sample when it is transferred to a 125 -m 1 . flask at the same temperature?

Indicate which represents the higher pressure in each of the following pairs: (a) 534 mm Hg or 0.754 bar (b) 534 mm Hg or 650 kPa (c) 1.34 bar or \(934 \mathrm{kPa}\)

Chloroform is a common liquid used in the laboratory. It vaporizes readily. If the pressure of chloroform vapori in a flask is \(195 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C}\) and the density of the vapor is \(1.25 \mathrm{g} / \mathrm{L},\) what is the molar mass of chloroform?

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