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We can get the average kinetic energy of a molecule, 1 2 mv 2 size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {} , from the left-hand side of the equation by canceling N size 12{N} {} and multiplying by 3/2. This calculation produces the result that the average kinetic energy of a molecule is directly related to absolute temperature.

KE ¯ = 1 2 m v 2 ¯ = 3 2 kT size 12{ {overline {"KE"}} = { {1} over {2} } m {overline {v rSup { size 8{2} } }} = { {3} over {2} } ital "kT"} {}

The average translational kinetic energy of a molecule, KE ¯ size 12{ {overline {"KE"}} } {} , is called thermal energy     . The equation KE ¯ = 1 2 m v 2 ¯ = 3 2 kT size 12{ {overline { size 11{"KE"}}} = { {1} over {2} } m {overline { size 11{v rSup { size 8{2} } }}} = { {3} over {2} } ital "kT"} {} is a molecular interpretation of temperature, and it has been found to be valid for gases and reasonably accurate in liquids and solids. It is another definition of temperature based on an expression of the molecular energy.

It is sometimes useful to rearrange KE ¯ = 1 2 m v 2 ¯ = 3 2 kT size 12{ {overline { size 11{"KE"}}} = { {1} over {2} } m {overline { size 11{v rSup { size 8{2} } }}} = { {3} over {2} } ital "kT"} {} , and solve for the average speed of molecules in a gas in terms of temperature,

v 2 ¯ = v rms = 3 kT m , size 12{ sqrt { {overline {v rSup { size 8{2} } }} } =v rSub { size 8{"rms"} } = sqrt { { {3 ital "kT"} over {m} } } ,} {}

where v rms size 12{v rSub { size 8{"rms"} } } {} stands for root-mean-square (rms) speed.

Calculating kinetic energy and speed of a gas molecule

(a) What is the average kinetic energy of a gas molecule at 20 . 0 º C size 12{"20" "." 0°C} {} (room temperature)? (b) Find the rms speed of a nitrogen molecule ( N 2 ) size 12{ \( N rSub { size 8{2} } \) } {} at this temperature.

Strategy for (a)

The known in the equation for the average kinetic energy is the temperature.

KE ¯ = 1 2 m v 2 ¯ = 3 2 kT size 12{ {overline {"KE"}} = { {1} over {2} } m {overline {v rSup { size 8{2} } }} = { {3} over {2} } ital "kT"} {}

Before substituting values into this equation, we must convert the given temperature to kelvins. This conversion gives T = ( 20 . 0 + 273 ) K = 293 K . size 12{T= \( "20" "." 0+"273" \) " K=293 K" "." } {}

Solution for (a)

The temperature alone is sufficient to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives

KE ¯ = 3 2 kT = 3 2 1 . 38 × 10 23 J/K 293 K = 6 . 07 × 10 21 J . size 12{ {overline {"KE"}} = { {3} over {2} } ital "kT"= { {3} over {2} } left (1 "." "38" times "10" rSup { size 8{ - "23"} } " J/K" right ) left ("293"" K" right )=6 "." "07" times "10" rSup { size 8{ - "21"} } `J "." } {}

Strategy for (b)

Finding the rms speed of a nitrogen molecule involves a straightforward calculation using the equation

v 2 ¯ = v rms = 3 kT m , size 12{ sqrt { {overline {v rSup { size 8{2} } }} } =v rSub { size 8{"rms"} } = sqrt { { {3 ital "kT"} over {m} } } ,} {}

but we must first find the mass of a nitrogen molecule. Using the molecular mass of nitrogen N 2 size 12{N rSub { size 8{2} } } {} from the periodic table,

m = 2 14 . 0067 × 10 3 kg/mol 6 . 02 × 10 23 mol 1 = 4 . 65 × 10 26 kg . size 12{m= { {2 left ("14" "." "0067" right ) times "10" rSup { size 8{ - 3} } `"kg/mol"} over {6 "." "02" times "10" rSup { size 8{"23"} } `"mol" rSup { size 8{ - 1} } } } =4 "." "65" times "10" rSup { size 8{ - "26"} } `"kg" "." } {}

Solution for (b)

Substituting this mass and the value for k size 12{k} {} into the equation for v rms size 12{v rSub { size 8{"rms"} } } {} yields

v rms = 3 kT m = 3 1 . 38 × 10 23 J/K 293 K 4 . 65 × 10 –26 kg = 511 m/s . size 12{v rSub { size 8{"rms"} } = sqrt { { {3 ital "kT"} over {m} } } = sqrt { { {3 left (1 "." "38" times "10" rSup { size 8{–"23"} } " J/K" right ) left ("293 K" right )} over {4 "." "65" times "10" rSup { size 8{"–26"} } " kg"} } } ="511"" m/s" "." } {}

Discussion

Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. The rms velocity of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule can move on average between collisions) of molecules in air is very small, and so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, however, which is about 340 m/s at room temperature. The faster the rms speed of air molecules, the faster that sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium. (See [link] .)

In part a of the figure, circles represent molecules distributed in a gas. Attached to each circle is a vector representing velocity. The circles have a random arrangement, while the vector arrows have random orientations and lengths. In part b of the figure, an arc represents a sound wave as it passes through a gas. The velocity of each molecule along the peak of the wave is roughly oriented parallel to the transmission direction of the wave.
(a) There are many molecules moving so fast in an ordinary gas that they collide a billion times every second. (b) Individual molecules do not move very far in a small amount of time, but disturbances like sound waves are transmitted at speeds related to the molecular speeds.

Questions & Answers

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Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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