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Making connections: historical note—kinetic theory of gases

The kinetic theory of gases was developed by Daniel Bernoulli (1700–1782), who is best known in physics for his work on fluid flow (hydrodynamics). Bernoulli’s work predates the atomistic view of matter established by Dalton.

Distribution of molecular speeds

The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This distribution is called the Maxwell-Boltzmann distribution , after its originators, who calculated it based on kinetic theory, and has since been confirmed experimentally. (See [link] .) The distribution has a long tail, because a few molecules may go several times the rms speed. The most probable speed v p size 12{v rSub { size 8{p} } } {} is less than the rms speed v rms size 12{v rSub { size 8{"rms"} } } {} . [link] shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds.

A line graph of probability versus velocity in meters per second of oxygen gas at 300 kelvin. The graph is skewed to the right, with a peak probability just under 400 meters per second and a root-mean-square probability of about 500 meters per second.
The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed v p size 12{v rSub { size 8{p} } } {} is less than the rms speed v rms size 12{v rSub { size 8{"rms"} } } {} . Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than v rms size 12{v rSub { size 8{"rms"} } } {} .

The distribution of thermal speeds depends strongly on temperature. As temperature increases, the speeds are shifted to higher values and the distribution is broadened.

Two distributions of probability versus velocity at two different temperatures plotted on the same graph. Temperature two is greater than Temperature one. The distribution for Temperature two has a peak with a lower probability, but a higher velocity than the distribution for Temperature one. The T sub two graph has a more normal distribution and is broader while the T sub one graph is more narrow and has a tail extending to the right.
The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures.

What is the implication of the change in distribution with temperature shown in [link] for humans? All other things being equal, if a person has a fever, he or she is likely to lose more water molecules, particularly from linings along moist cavities such as the lungs and mouth, creating a dry sensation in the mouth.

Calculating temperature: escape velocity of helium atoms

In order to escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity . At what temperature would helium atoms have an rms speed equal to the escape velocity?

Strategy

Identify the knowns and unknowns and determine which equations to use to solve the problem.

Solution

1. Identify the knowns: v size 12{v} {} is the escape velocity, 11.1 km/s.

2. Identify the unknowns: We need to solve for temperature, T size 12{T} {} . We also need to solve for the mass m size 12{m} {} of the helium atom.

3. Determine which equations are needed.

  • To solve for mass m size 12{m} {} of the helium atom, we can use information from the periodic table:
    m = molar mass number of atoms per mole . size 12{m= { { size 11{"molar mass"}} over { size 11{"number of atoms per mole"}} } } {}
  • To solve for temperature T size 12{T} {} , we can rearrange either
    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"} {}

    or

    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} } } } {}

    to yield

    T = m v 2 ¯ 3 k , size 12{T= { {m {overline {v rSup { size 8{2} } }} } over {3k} } ,} {}
    where k size 12{k} {} is the Boltzmann constant and m size 12{m} {} is the mass of a helium atom.

4. Plug the known values into the equations and solve for the unknowns.

m = molar mass number of atoms per mole = 4 . 0026 × 10 3 kg/mol 6 . 02 × 10 23 mol = 6 . 65 × 10 27 kg size 12{m= { { size 11{"molar mass"}} over { size 11{"number of atoms per mole"}} } = { { size 11{4 "." "0026" times "10" rSup { size 8{ - 3} } " kg/mol"}} over { size 12{6 "." "02" times "10" rSup { size 8{"23"} } " mol"} } } =6 "." "65" times "10" rSup { size 8{ - "27"} } " kg"} {}
T = 6 . 65 × 10 27 kg 11 . 1 × 10 3 m/s 2 3 1 . 38 × 10 23 J/K = 1 . 98 × 10 4 K size 12{T= { { left (6 "." "65" times "10" rSup { size 8{ - "27"} } `"kg" right ) left ("11" "." 1 times "10" rSup { size 8{3} } `"m/s" right ) rSup { size 8{2} } } over {3 left (1 "." "38" times "10" rSup { size 8{ - "23"} } `"J/K" right )} } =1 "." "98" times "10" rSup { size 8{4} } `K} {}

Discussion

This temperature is much higher than atmospheric temperature, which is approximately 250 K ( 25 º C size 12{ \( –"25"°C} {} or 10 º F ) size 12{–"10"°F \) } {} at high altitude. Very few helium atoms are left in the atmosphere, but there were many when the atmosphere was formed. The reason for the loss of helium atoms is that there are a small number of helium atoms with speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one instant to the next, so that at any instant, there is a small, but nonzero chance that the speed is greater than the escape speed and the molecule escapes from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water (very little of which reach a very high altitude), have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, so few have speeds above the escape velocity that billions of years are required to lose significant amounts of the atmosphere. [link] shows the impact of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The comparison between Earth and the Moon is discussed in this chapter’s Problems and Exercises.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
what is viscosity?
David
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emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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Mohammed
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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