# Kinetic Theory of Gases: Molecular Speeds

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 (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

The quantity *u*_{rms} 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,

**KE**where 1 and 2 denote the two gases. Since KE= 1/2mv_{1}= KE_{2}^{2},

**1/2 m**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_{1}(*u*_{rms, }_{1})^{2}= 1/2 m_{2}(*u*_{rms, }_{2})^{2}^{-24}g./average molecule, and*u*_{rms}is the velocity.

Dividing,

**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 H_{2} 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.**

**m**_{2}= 121

**EXAMPLE 2** Find the rms velocity for (a) H_{2} and (b) O_{2} 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 m^{2}s^{–2} mol^{–1} K^{–1}

**a)** For H_{2}

**b)** For O_{2}

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 O_{2} molecules in air at room temperature move about 50 percent faster than jet planes, and H_{2} molecules are nearly 4 times speedier yet. Of course an O_{2} 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 H_{2}, O_{2} 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 H_{2} or O_{2} 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 H_{2} and O_{2} molecules occupy separate containers at the same temperature and pressure. The volumes are seen to be the same. Because O_{2} molecules are 16 times heavier than H_{2} molecules, the average speed of H_{2} molecules is 4 times faster. H_{2} molecules therefore make 4 times as many collisions with walls. Based on mass, each collision of an H_{2} molecule with the wall has one-sixteenth the effect of an O_{2} collision, but an H_{2} collision has 4 times the effect of an O_{2} collision when molecular velocity is considered. The net result is that each H_{2} collision is only one-fourth as effective as an O_{2} collision.

Four times as many collisions, each one-fourth as effective, results in the same pressure. Thus the same number of O_{2} molecules as H_{2} molecules is required to occupy the same volume at the same temperature and pressure.

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