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\(\mathrm{WF}_{6}\) is one of the heaviest known gases. How much slower is the root-mean-square speed of \(\mathrm{WF}_{6}\) than He at \(300 \mathrm{~K}\) ?

Short Answer

Expert verified
The root-mean-square speed of $\mathrm{WF}_6$ is 1154.02 m/s slower than that of He at 300 K.

Step by step solution

01

Find the molecular masses of WF6 and He

To determine the RMS speed, we first need to find the molecular masses of WF6 and helium. The molecular mass of a compound or element is obtained by multiplying the number of atoms of each element by its atomic mass. The molecular mass of WF6 and He are : Molecular mass of WF6 = (1 * atomic mass of W) + (6 * atomic mass of F) Molecular mass of He = atomic mass of He Find the atomic masses of W, F, and He from the periodic table: Atomic mass of W = 183.84 u Atomic mass of F = 18.998 u Atomic mass of He = 4.0026 u Calculate the molecular masses: Molecular mass of WF6 = (1 * 183.84) + (6 * 18.998) = 183.84 + 113.988 = 297.828 u Molecular mass of He = 4.0026 u Now, we need to convert these molecular masses to their SI units (kg/mol): 1 u = 1.66054 x 10^(-27) kg Mass of WF6 = 297.828 u * 1.66054 x 10^(-27) kg/u = 4.9471 x 10^(-25) kg Mass of He = 4.0026 u * 1.66054 x 10^(-27) kg/u = 6.64665 x 10^(-27) kg
02

Calculate the RMS speeds of WF6 and He

We will now use the RMS speed formula to calculate the speeds of WF6 and He: \(v_{rms} = \sqrt{\frac{3k_bT}{m}}\) Where Boltzmann's constant (\(k_b\)) = 1.38064852 × 10^(-23) m² kg s² K^(-1) and the temperature (T) = 300 K. RMS speed of WF6: \(v_{rms}(WF_6) = \sqrt{\frac{3(1.38064852 × 10^{-23} \mathrm{m^2\cdot kg\cdot s^{-2} \cdot K^{-1}})(300 \mathrm{K})}{(4.9471 × 10^{-25} \mathrm{kg})}} = 215.73 \, \mathrm{m/s}\) RMS speed of He: \(v_{rms}(He) = \sqrt{\frac{3(1.38064852 × 10^{-23} \mathrm{m^2\cdot kg\cdot s^{-2} \cdot K^{-1}})(300 \mathrm{K})}{(6.64665 × 10^{-27} \mathrm{kg})}} = 1369.75 \, \mathrm{m/s}\)
03

Calculate the difference in speeds

Now we can find the difference in RMS speeds between WF6 and He: Difference = RMS speed of He - RMS speed of WF6 Difference = 1369.75 m/s - 215.73 m/s = 1154.02 m/s The root-mean-square speed of WF6 is 1154.02 m/s slower than that of He at 300 K.

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Key Concepts

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

Molecular Mass

The concept of molecular mass is pivotal in determining various properties of gases, including their speed, as shown in the provided exercise. Molecular mass, often expressed in unified atomic mass units (u), reflects the total weight of all the atoms present in a molecule. In our example with WF6 and He, the molecular mass is calculated by considering the atomic mass of individual elements obtained from the periodic table.

  • For WF6, the molecular mass is the sum of the mass of one tungsten (W) atom and six fluorine (F) atoms.
  • For He, the molecular mass is simply the mass of one helium atom.

Since molecular mass is a crucial factor in determining the root-mean-square speed of a gas, as shown in the RMS speed formula, understanding it allows us to compare the kinetic behaviours of different gases under the same conditions.

Gas Kinetics

In the study of gas kinetics, we analyze the motion and behaviour of gas particles. The root-mean-square (RMS) speed is a direct application of gas kinetics, giving a measure of the average speed of particles in a gas sample. It's derived from the kinetic theory, which determines the speed based on the kinetic energy associated with the temperature of the gas and the mass of its particles.


Understanding RMS Speed

The RMS speed takes into account the distribution of speeds among gas molecules, offering a middle ground that defines the overall kinetic energy in the system. Our exercise utilized the RMS speed to establish a comparison between WF6 and He.

Higher molecular mass, as seen with WF6, generally results in a slower RMS speed, given the same amount of kinetic energy derived from the same temperature. This concept allows us to anticipate how gases diffuse and react, playing a critical role in fields like meteorology and materials science.

Boltzmann's Constant

Boltzmann's constant (\(k_b\)) is a fundamental physical constant that occurs in various equations in physics, particularly within the realm of thermodynamics and kinetic theory. It relates the average kinetic energy per particle in a gas with the temperature of the gas, acting as a bridge between microscopic and macroscopic views of thermodynamics.

  • It has a value of approximately 1.38064852 × 10^(-23) m² kg s² K^(-1).
  • It's prominently featured in the RMS speed formula used in our exercise to quantify the motion of gas particles at a given temperature (\(T\)).

Through Boltzmann's constant, we can understand that even if we cannot measure the energy or speed of individual molecules, we can calculate their average properties that emerge from their collective behaviour — a striking feature that's central to the study of thermodynamics and statistical mechanics.

Temperature in Kinetic Theory

In kinetic theory, temperature is not merely a measure of 'hotness' or 'coldness', but rather an indication of the average kinetic energy of particles in a substance. It is directly proportional to this energy, meaning that as the temperature of a gas increases, so does the kinetic energy (and hence the speed) of its particles.


Role of Temperature in Gas Kinetics

Temperature is usually measured in Kelvins (K), where an increase in Kelvin correlates to an increase in particle speed. In the exercise we analyzed, both WF6 and He were at 300 K, allowing for a fair comparison of their RMS speeds. Importantly, this means that the massive difference in the RMS speeds of the two gases at the same temperature can be attributed to the difference in their molecular masses rather than any difference in thermal energy content.

Understanding this aspect of temperature is crucial for predicting and explaining the behaviour of gases in various practical and theoretical scenarios, from the workings of an engine to the diffusion of gases in the atmosphere.

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Most popular questions from this chapter

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