According to the kinetic-molecular theory of gases, the average kinetic energy of any collection of gas molecules, m(urms)2, has a specific value at a given temperature. Thus, for two gases at the same temperature a gas composed of low-mass particles, such as He, has the same average kinetic energy as one composed of more massive particles, such as Xe. The mass of the particles in the He sample is smaller than that in the Xe sample. Consequently, the He particles must have a higher rms speed than the Xe particles. The equation that expresses this fact quantitatively is

where M is the molar mass of the particles, which can be derived from the kinetic-molecular theory. Because M appears in the denominator, the less massive the gas particles, the higher their rms speed.


How does root-mean-square speed vary with molar mass?

FIGURE 10.18 The effect of molar mass on molecular speed at 25 °C.

FIGURE 10.18 shows the distribution of molecular speeds for several gases at 25 °C. Notice how the distributions are shifted toward higher speeds for gases of lower molar masses.

SAMPLE EXERCISE 10.14 Calculating a Root-Mean-Square Speed

Calculate the rms speed of the molecules in a sample of N2 gas at 25 °C.


Analyze We are given the identity of a gas and the temperature, the two quantities we need to calculate the rms speed.

Plan We calculate the rms speed using Equation 10.22.

Solve We must convert each quantity in our equation to SI units. We will also use R in units of J/mol-K (Table 10.2) to make the units cancel correctly.

Comment This corresponds to a speed of 1150 mi/hr. Because the average molecular weight of air molecules is slightly greater than that of N2, the rms speed of air molecules is a little lower than that for N2.


What is the rms speed of an atom in a sample of He gas at 25 °C?

Answer: 1.36 × 103 m/s

The most probable speed of a gas molecule can also be derived:


What is the ratio of urms to ump for a certain gas at a given temperature?

FIGURE 10.19 Effusion.

The dependence of molecular speed on mass has two interesting consequences. The first is effusion, which is the escape of gas molecules through a tiny hole (FIGURE 10.19). The second is diffusion, which is the spread of one substance throughout a space or throughout a second substance. For example, the molecules of a perfume diffuse throughout a room.

Graham's Law of Effusion

In 1846 Thomas Graham (1805–1869) discovered that the effusion rate of a gas is inversely proportional to the square root of its molar mass. Assume we have two gases at the same temperature and pressure in two containers with identical pinholes. If the rates of effusion of the two gases arer1 and r2 and their molar masses are M1 and M2, Graham's law states that

a relationship that indicates that the lighter gas has the higher effusion rate.

The only way for a molecule to escape from its container is for it to “hit” the hole in the partitioning wall of Figure 10.19. The faster the molecules are moving, the greater is the likelihood that a molecule will hit the hole and effuse. This implies that the rate of effusion is directly proportional to the rms speed of the molecules. Because R and T are constant, we have, from Equation 10.22

As expected from Graham's law, helium escapes from containers through tiny pin-hole leaks more rapidly than other gases of higher molecular weight (FIGURE 10.20).


Because pressure and temperature are constant in this figure but volume changes, which other quantity in the ideal-gas equation must also change?

FIGURE 10.20 An illustration of Graham's law of effusion.

SAMPLE EXERCISE 10.15 Applying Graham's Law

An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is 0.355 times the rate at which O2 gas effuses at the same temperature. Calculate the molar mass of the unknown and identify it.


Analyze We are given the rate of effusion of an unknown gas relative to that of O2 and asked to find the molar mass and identity of the unknown. Thus, we need to connect relative rates of effusion to relative molar masses.

Plan We use Equation 10.24, to determine the molar mass of the unknown gas. If we let rx and Mx represent the rate of effusion and molar mass of the gas, we can write

Solve From the information given,

rx = 0.355 × rO2


Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.


Calculate the ratio of the effusion rates of N2 gas and O2 gas.

Answer: rN2/rO2 = 1.07

Diffusion and Mean Free Path

Although diffusion, like effusion, is faster for lower-mass molecules than for higher-mass ones, molecular collisions make diffusion more complicated than effusion.

Graham's law, Equation 10.24, approximates the ratio of the diffusion rates of two gases under identical conditions. We can see from the horizontal axis in Figure 10.18 that the speeds of molecules are quite high. For example, the rms speed of molecules of N2 gas at room temperature is 515 m/s. In spite of this high speed, if someone opens a vial of perfume at one end of a room, some time elapses—perhaps a few minutes—before the scent is detected at the other end of the room. This tells us that the diffusion rate of gases throughout a volume of space is much slower than molecular speeds.* This difference is due to molecular collisions, which occur frequently for a gas at atmospheric pressure—about 1010 times per second for each molecule. Collisions occur because real gas molecules have finite volumes.

Gas Separations

The fact that lighter molecules move at higher average speeds than more massive ones has many interesting applications. For example, developing the atomic bomb during World War II required scientists to separate the relatively low-abundance uranium isotope 235U (0.7%) from the much more abundant 238U (99.3%). This separation was accomplished by converting the uranium into a volatile compound, UF6, that was then allowed to pass through a porous barrier. (Because of the pore diameters, this process is not simple effusion. Nevertheless, the way in which rate of passing through the pores depends on molar mass is essentially the same as in effusion.) The slight difference in molar mass between 235UF6 and 238UF6 caused the molecules to move at slightly different rates:

Thus, the gas initially appearing on the opposite side of the barrier was very slightly enriched in 235U. The process was repeated thousands of times, leading to a nearly complete separation of the two isotopes.

Separation of uranium isotopes by effusion has been largely replaced by a technique that uses centrifuges. In this procedure, cylindrical rotors containing UF6 vapor spin at high speed inside an evacuated casing. Molecules of 235UF6 move closer to the spinning walls, whereas molecules of 235UF6 remain in the middle of the cylinders. A stream of gas moves the 235UF6 from the center of one centrifuge into another. Plants that use centrifuges consume less energy than those that use effusion and can be constructed in a more compact, modular fashion. Such plants are frequently in the news today as countries such as Iran and North Korea enrich uranium in the 235U isotope for both nuclear power and nuclear weaponry.

RELATED EXERCISES: 10.89 and 10.90

Because of molecular collisions, the direction of motion of a gas molecule is constantly changing. Therefore, the diffusion of a molecule from one point to another consists of many short, straight-line segments as collisions buffet it around in random directions (FIGURE 10.21).

The average distance traveled by a molecule between collisions, called the molecule's mean free path, varies with pressure as the following analogy illustrates. Imagine walking through a shopping mall. When the mall is crowded (high pressure), the average distance you can walk before bumping into someone is short (short mean free path). When the mall is empty (low pressure), you can walk a long way (long mean free path) before bumping into someone. The mean free path for air molecules at sea level is about 60 nm. At about 100 km in altitude, where the air density is much lower, the mean free path is about 10 cm, over 1 million times longer than at Earth's surface.


Will these changes increase, decrease, or have no effect on the mean free path of the molecules in a gas sample?

a. increasing pressure.

b. increasing temperature.

FIGURE 10.21 Diffusion of a gas molecule. For clarity, no other gas molecules in the container are shown.