Skip to main content
Chemistry LibreTexts

Solutions 1

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)


    Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:

    \[E_{rms} = \frac{3VP}{2N} (1)\]

    1. The pressure of the gas is increased by reducing the volume at constant temperature.
    2. The pressure of the gas is increased by increasing the temperature at constant volume.
    3. The average velocity of the molecules is increased by a factor of 2.

    a) If the pressure is increased with the volume being decreased in equal measure, from equation 1 we can see the average kinetic energy remains unchanged.

    b) Given the ideal gas law PV = nRT if T goes up and V is constant then the pressure P must increase. If P increases then by equation 1 the average kinetic increases.

    c) the kinetic energy is related to the velocity by this equation. \[E_{rms} = \frac{1}{2}mv^2_{rms}\] so if the velocity doubles the kinetic energy increases by four.


    What assumptions are made when applying the kinetic molecular theory to gases? Are all of these assumptions necessary? Why or why not?

    The assumptions are:

    1. The gas consists of objects with a defined mass and zero volume.

    2. Gas particles move in straight lines unless acted upon by collision with another particles or container walls.

    3. All collisions between gas particles and each other and the container walls are elastic

    4. Gas particles do not interact aside from colliding with each other.

    5. The average kinetic energy of the gas is proportional to the temperature.

    These assumptions allow us to approximately calculate the macroscopic properties of a gas without dealing with a lot of complications.


    The root-mean-square speed of 6 particles is 2.47 ms-1. The speed of 5 of the particles are 1.0, 2, 1.5, 3.0 and 2.5. Calculate the unknown speed of the 6th particle. Find the average speed of the 6 particles.

    \[V^2_{rms} = \frac{V^2_{1} + V^2_{2} + V^2_{3} + ... + V^2_{N}}{N}\]

    \[6.1009 = \frac{1 + 4 + 2.25 + 9 + 6.25 + V^2_{6}}{6}\]

    \[V^2_{6} = 36.6 - 22.5 = 14.1054\]

    \[V_{6} = 3.76\]

    The average speed of the particles is just equal to all the speeds added together and divided by six.

    \[V_{avg} = \frac{1 + 2 + 1.5 + 3 + 2.5 + 3.76}{6} = 2.29\]


    Consider the distribution of molecular velocities in a sample of helium. If the sample is cooled, will the distribution of velocities look more like that of H2 or of H2O? Explain your answer.

    \[V_{rms} = \sqrt{\frac{3RT}{M}}\]

    It should look more like water because as far as velocity goes decreasing the temperature could be modelled equally as well by increasing the mass of the gas particles.


    Find the \(v_{rms}\) of N_{2(g)} at 25ºC. What temperature must \(Cl_{2(g)}\) be to have the same \(v_{rms}\)?

    \[v^2_{rms} = \frac{3RT}{M}\]

    For nitrogen with at 298 K with a molar mass of .028 kg/mol

    \[v^2_{rms} = 265454 \frac{m^2}{s^2}\]

    \[v_{rms} = 515.22 \frac{m}{s}\]

    For chlorine with a molar mass of .0709 kg/mol the temperature at which its root-mean-square velocity is equal to nitrogen's at 298 K is

    \[T = \frac{v^2_{rms}*T}{3R}\]

    \[T = \frac{265454 * .0709}{24.942} = 754.5 K \]


    Use the Maxwell formula

    \[f(v)=4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

    to derive an expression for \(c_{mp}\) of a gas.

    \(c_{mp}\) is determined by finding the value of v where the derivative of f(v) with respect to v is zero.

    \[\frac{df(v)}{dv} = 0 = 8\pi v \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] - (\frac{mv}{k_bT})4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

    Then rearranging the terms we have:

    \[8\pi v \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] = (\frac{mv}{k_bT})4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

    Cancelling out terms on both sides

    \[2 = v^2 * (\frac{m}{k_bT})\]

    \[v_{mp} = \sqrt{\frac{2k_bT}{m}} = \sqrt{\frac{2RT}{M}} = c_{mp}\]

    Solutions 1 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?