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Kinetic Theory of Gases Equation (OCR A Level Physics)

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Katie M

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Katie M

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The Root Mean Square Speed

  • The kinetic theory of gases equation includes the mean square speed of the particles:

top enclose bold c to the power of bold 2 end enclose

  • Where

    • c = average speed of the gas particles

    • top enclose c squared end enclose has the units m2 s−2 

  • Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction

  • When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall

  • In order to find the pressure of the gas, the velocities must be squared

    • This is a more useful method, since a negative or positive number squared is always positive

  • To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

square root of stack bold c to the power of bold 2 with bold bar on top end root

  • This is known as the root-mean-square speed and still has the units of m s−1

    • The root-mean-square speed can also have the symbol cr.m.s.

  • The mean square speed is not the same as the mean speed

Worked Example

A very small group of atoms have the following velocities: 

50 m s−1

+80 m s−1

+85 m s−1

−65 m s−1

−90 m s−1

Calculate the mean speed, c with bar on top, mean square speed, stack c squared with bar on top, and r.m.s speed, cr.m.s, of these atoms.

Answer:

Step 1: Calculate the mean speed c with bar on top:

  • Add all the values and divide by the number of values you have to calculate the mean

c with bar on top = fraction numerator 50 space plus space 80 space plus 85 space minus 65 space minus 90 over denominator 5 end fraction = +12 m s−1 

Step 2: Calculate the mean square speed stack c squared with bar on top:

  • Square each value, then add them all and divide by the number of values you have to calculate the mean of the squares

stack c squared with bar on top = fraction numerator 50 squared plus 80 squared plus 85 squared space plus left parenthesis negative 65 right parenthesis squared plus left parenthesis negative 90 right parenthesis squared over denominator 5 end fraction = +5690 m2 s−2 

  • Here the units are also squared, as (m s−1)2 = m2 s−2 

Step 3: Calculate the r.m.s speed cr.m.s

  • Find the square root of the mean square speed stack c squared with bar on top

cr.m.s  = square root of stack c squared with bar on top end root = square root of 5690 space straight m squared straight s to the power of negative 2 end exponent end root space = 75.4 m s−1 = 75 m s−1 (2 s.f.) 

Examiner Tips and Tricks

Make sure you read questions relating to the r.m.s speed carefully! It is easy to get confused between c with bar on top, stack c squared with bar on top and square root of stack c squared with bar on top end root

Kinetic Theory of Gases Equation

  • The Kinetic Theory of Gases Equation is given by:

bold italic p bold italic V bold equals bold 1 over bold 3 bold italic N bold space bold italic m bold space stack bold c to the power of bold 2 with bold bar on top

  • Where 

    • p = pressure (Pa)

    • V = volume (m3)

    • N = number of molecules

    • m = mass of one molecule of gas (kg)

    • stack c squared with bar on top = mean square speed of the molecules (m2 s−2)

  • The density, ρ of the gas is given by:

rho equals mass over volume equals fraction numerator N m over denominator V end fraction

  • Rearranging the equation for pressure, p and substituting the density, ρ gives the equation:

p equals 1 third space rho space stack c squared with bar on top

  • Where 

    • p = pressure (Pa)

    • ρ = density (kg m−3)

    • stack c squared with bar on top = mean square speed of the molecules (m2 s−2)

Worked Example

A gas cylinder contains argon at a pressure of 700 kPa.

The cylinder contains 3.0 × 1025 molecules and each molecule has a mass of 4.7 × 10−26 kg. The r.m.s speed of the molecules is 350 m s−1.

Calculate the volume of the cylinder. 

Answer:

Step 1: List the known quantities

  • Number of molecules, N = 3.0 × 1025

    • Mass of each molecule, m = 4.7 × 10−26 kg

    • Root mean square speed, cr.m.s. = 350 m s−1

      • Therefore stack c squared with bar on top = 3502 m2s−2

    • Pressure, P = 700 kPa = 700 × 103 Pa

Step 2: State the equation

p V equals 1 third space N space m space stack straight c squared with bar on top

Step 3: Rearrange the equation to make volume, V, the subject

V equals fraction numerator N m stack c squared with bar on top over denominator 3 p end fraction

Step 3: Substitute the values from the question into the equation and calculate V

V equals fraction numerator left parenthesis 3.0 space cross times space 10 to the power of 25 right parenthesis space cross times space left parenthesis 4.7 space cross times space 10 to the power of negative 26 end exponent right parenthesis space cross times space left parenthesis 350 right parenthesis squared over denominator 3 space cross times space left parenthesis 700 space cross times space 10 cubed right parenthesis end fraction

V = 0.08225 m3

Step 4: Write the final answer to the correct number of decimal places

  • The volume of the cylinder is 0.082 m3 (3 d.p.)

Worked Example

An ideal gas has a density of 4.5 kg m−3 at a pressure of 9.3 × 105 Pa.

Determine the root mean square speed, cr.m.s., of the gas atoms at a constant temperature.

Answer:

Step 1: Write out the known quantities

  • Pressure, p = 9.3 × 105 Pa

  • Density, ρ = 4.5 kg m−3

Step 2: State the equation linking the pressure of an ideal gas with density

p equals 1 third rho stack c squared with bar on top

Step 3: Rearrange to make stack bold italic c to the power of bold 2 with bar on top the subject

stack c squared with bar on top equals fraction numerator 3 p over denominator rho end fraction

Step 4: Take square roots of both sides to give an equation for cr.m.s.

c subscript r. m. s. end subscript equals square root of fraction numerator 3 p over denominator rho end fraction end root

Step 5: Substitute in known values and calculate cr.m.s.

c subscript r. m. s. end subscript equals square root of fraction numerator 3 cross times 9.3 cross times 10 to the power of 5 over denominator 4.5 end fraction end root equals 787.4 space equals space 790 space straight m space straight s to the power of negative 1 end exponent

Examiner Tips and Tricks

You are not expected to know the derivation of the kinetic theory equation. However, a common exam question is to use the kinetic model of a gas and Newton's laws of motion to explain how a gas exerts pressure on the walls of its container.

You will also be expected to explain why the mean square speed of the molecules is used in this equation instead of mean velocity. It is important to note that these quantities are not equivalent. The mean velocity would be zero because particles with opposite velocities cancel out. 

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.