Kinetic Theory of Gases: Molecular Speeds

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:27

Other sections state that increasing the temperatureA physical property that indicates whether one object can transfer thermal energy to another object. increases the speeds at which molecules move. We are now in a position to find just how large that increase is for a gaseous substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula.. Combining the ideal gasA hypothetical gas for which the relationship among the pressure, volume, temperature, and chemical amount (moles) can be described by simple proportionalities summarized by the ideal gas equation, PV = nRT. law with Eq. (1) from The Total Molecular Kinetic Energy, we obtain


or       \text{3}RT=\frac{Nm}{n}\text{(}u^{\text{2}}\text{)}_{\text{ave}} (1)

Since N is the number of molecules and m is the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of each molecule, Nm is the total mass of gas. Dividing total mass by amount of substance gives molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. M:


Substituting in Eq. (1), we have

  & \text{                                       3}RT=M(u^{\text{2}})_{\text{ave}} \ 
 & \text{or                                   }(u^{\text{2}})_{\text{ave}}=\frac{\text{3}RT}{M} \ 
 & \text{so that                               }u_{rms}=\sqrt{\text{(}u^{\text{2}}\text{)}_{\text{ave}}}=\sqrt{\frac{\text{3}RT}{M}}\text{                                                                    (2)} \ 

Figure 1 A microscopic interpretation of Gay-Lussac’s law. As the temperature of a gas is increased, the velocity of the molecules is also increased. More molecules hit the sides of the container, each with a greater impulse, so that the pressure increases.

The quantity urms is called the root-mean-square (rms) velocity because it is the square root of the mean square velocity.

The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Thus quadrupling the temperature of a given gas doubles the rms velocity of the molecules. Doubling this average velocity doubles the number of collisions between gas molecules and the walls of a container. It also doubles the impulse of each collision. Thus the pressure quadruples. This is indicated graphically in Fig. 1. Pressure is thus directly proportional to temperature, as required by Gay-Lussac’s law.

The inverse proportionality between root-mean-square velocity and the square root of molar mass means that the heavier a molecule is, the slower it moves. This is illustrated in the examples below.

We can compare the rates of effusionThe escape of a gas through an orifice. or diffusionThe spreading of one substance into another (usually involves gases or liquids). of a known gas with that of an unknown gas to determine the molar mass of the unknown gas. A convenient equation can be derived easily by considering the kinetic energy of individual molecules rather than moles of gas:

Knowing that kinetic energy is proportional to temperature, if the two gases are at the same temperature,

KE1 = KE2   where 1 and 2 denote the two gases. Since KE= 1/2mv2,
1/2 m1 (urms,  1)2 = 1/2 m2 (urms,  2)2   where m is the atomic weightThe average mass of the naturally occurring isotopes of an element, taking into account the different natural abundances of the isotopes. Expressed relative to the value of exactly 12 for carbon-12; also called atomic mass. in amuAbbreviation for atomic mass unit, a unit for expressing the relative masses of atoms. An atom of carbon-12 has a mass of 12 amu. 1 amu corresponds to 1.662x10-24 g./average molecule, and urms is the velocity.


\frac{\text{m}_1}{\text{m}_2}  = \frac{u_{rms, ~ 2}^{2}}{u_{rms, ~ 1}^2}

EXAMPLE 1 What is the molar mass of an unknown gas if the gas effuses through a pinhole into a vacuum at a rate of 2 mL/min, and H2 effuses at 11 mL/min. Assume that the rate of effusion is proportional to the gas molecule velocities.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.

\frac{\text{m}_1}{\text{m}_2}  = \frac{u_{rms, ~ 2}^{2}}{u_{rms, ~ 1}^2}
\frac{4}{\text{m}_2}  = \frac{2^{2}}{11^2}
m2 = 121

EXAMPLE 2 Find the rms velocity for (a) H2 and (b) O2 molecules at 27°C.

Solution This problem is much easier to solve if we use SI unitsThe international system of units (Système International d'Unité) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole.. Thus we choose

R = 8.314 J mol–1 K–1 = 8.314 kg m2s–2 mol–1 K–1

a) For H2

\begin{align}u_{\text{rms}}=\sqrt{\frac{\text{3}RT}{M}}&=\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}{\text{2}\text{.016 g mol}^{-\text{1}}}}\
&=\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{kg m}^{\text{2}}\text{s}^{-\text{2}}}{\text{g}}}\
&=\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{g m}^{\text{2}}\text{s}^{-\text{2}}}{\text{g}}}\
&=\sqrt{\text{3}\text{.712  }}\times \text{ 10}^{\text{3}}\text{ m s}^{-\text{1}}=\text{1}\text{.927 }\times \text{ 10}^{\text{3}}\text{ m s}^{-\text{1}}\end{align}

b) For O2

u_{\text{rms}}=\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}{\text{32}\text{.00 g mol}^{-\text{1}}}}=\text{4}\text{.836 }\times \text{ 10}^{\text{2}}\text{ m s}^{-\text{1}}

Figure 2 Avogadro’s law. Equal number of (a) O2 and (b) H2 molecules are shown in separate containers at the same P. The speedier H2 molecules make 4 times as many collisions with the walls, but each collision by one of the heavier O2 molecules is 4 times as effective. Therefore both gases push the piston up to the same height and occupy the same volume.

The rms velocities 1927 m s–1 and 484 m s–1 correspond to about 4300 miles per hour and 1080 miles per hour, respectively. The O2 molecules in air at room temperature move about 50 percent faster than jet planes, and H2 molecules are nearly 4 times speedier yet. Of course an O2 molecule would take a lot longer to get from New York to Chicago than a jet would. Gas molecules never go far in a straight line before colliding with other molecules. Now we can see the microscopic basis for Avogadro’s law. Most of the volume in H2, O2 or any gas is empty space, and that empty space is the same for a given amount of any gas at the same temperature and pressure. This happens because the total kinetic energy of the molecules is the same for H2 or O2 or any other gas. The more energy they have, the more room the molecules can make for themselves by expanding against a constant pressure. This is illustrated in Fig. 2, where equal numbers of H2 and O2 molecules occupy separate containers at the same temperature and pressure. The volumes are seen to be the same. Because O2 molecules are 16 times heavier than H2 molecules, the average speed of H2 molecules is 4 times faster. H2 molecules therefore make 4 times as many collisions with walls. Based on mass, each collision of an H2 molecule with the wall has one-sixteenth the effect of an O2 collision, but an H2 collision has 4 times the effect of an O2 collision when molecular velocity is considered. The net result is that each H2 collision is only one-fourth as effective as an O2 collision.

Four times as many collisions, each one-fourth as effective, results in the same pressure. Thus the same number of O2 molecules as H2 molecules is required to occupy the same volume at the same temperature and pressure.