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Molecular energy, temperature and r.m.s. speed

Key idea: An ideal gas molecule's mean translational kinetic energy is 3kT/2. At the same kelvin temperature, heavier molecules have the same mean energy but a lower r.m.s. speed.

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  • GCE A-Level H2 Physics 2027
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The kinetic theory gives the pressure of a gas in terms of its molecules’ speeds; the equation of state gives the same pressure in terms of its temperature. Putting the two together reveals what temperature actually measures: the average translational kinetic energy of the molecules.

Temperature and the mean kinetic energy of a molecule

For the same sample of an ideal gas, pV = (1/3)Nm⟨c²⟩ and pV = NkT. Equating them and cancelling N:

(1/3)m⟨c²⟩ = kT

Multiplying both sides by 3/2,

(1/2)m⟨c²⟩ = (3/2)kT

The left-hand side is the mean translational kinetic energy of a molecule: the average kinetic energy due to the motion of its centre of mass. So the mean translational kinetic energy of a molecule of an ideal gas is directly proportional to its thermodynamic temperature. Double the temperature in kelvin, and the mean kinetic energy doubles.

Two points to notice:

  • It is a mean. At any instant the molecules have a wide spread of speeds and energies; (3/2)kT is their average.
  • The mass of the molecule does not appear in (3/2)kT. At the same temperature, the molecules of every ideal gas have the same mean translational kinetic energy, whether they are light helium atoms or heavy carbon dioxide molecules.

Mean translational energy against temperature

A straight line through the origin has energy 6.21 times ten to the minus 21 joules at 300 kelvin.

Scroll across the figure to read all labels.

A straight line through the origin has energy 6.21 times ten to the minus 21 joules at 300 kelvin.A straight line through the origin has energy 6.21 times ten to the minus 21 joules at 300 kelvin.
Calculated ⟨Eₜᵣₐₙₛ⟩ = 3kT/2 with k = 1.38 × 10⁻²³ J K⁻¹. The origin is the ideal-model limit, not an assertion that a real gas remains gaseous at 0 K.
Open full-size graph
What to noticeAt any fixed temperature, this mean translational energy is independent of molecular mass.
View figure data
Figure data: Mean translational energy against temperature
Temperature, T (K)Any ideal-gas molecular mass
00.000
100.207
200.414
300.621
400.828
501.035
601.242
701.449
801.656
901.863
1002.070
1102.277
1202.484
1302.691
1402.898
1503.105
1603.312
1703.519
1803.726
1903.933
2004.140
2104.347
2204.554
2304.761
2404.968
2505.175
2605.382
2705.589
2805.796
2906.003
3006.210
3106.417
3206.624
3306.831
3407.038
3507.245
3607.452
3707.659
3807.866
3908.073
4008.280
4108.487
4208.694
4308.901
4409.108
4509.315
4609.522
4709.729
4809.936
49010.143
50010.350
51010.557
52010.764
53010.971
54011.178
55011.385
56011.592
57011.799
58012.006
59012.213
60012.420
61012.627
62012.834
63013.041
64013.248
65013.455
66013.662
67013.869
68014.076
69014.283
70014.490

Root-mean-square speed

The root-mean-square speed is the square root of the mean of the squares of the speeds:

c_(r.m.s.) = square root of (⟨c²⟩)

Square each speed, find the mean, then take the square root, in that order. For four molecules moving at 100, 100, 300 and 300 m s⁻¹, the mean speed is 200 m s⁻¹, but the mean of the squares is 50 000 m² s⁻², so c_(r.m.s.) = 224 m s⁻¹. The r.m.s. speed is the one that goes with kinetic energy.

Rearranging (1/2)m⟨c²⟩ = (3/2)kT,

c_(r.m.s.) = square root of (3kT/m) = square root of (3RT/M)

using either the mass m of one molecule with k, or the molar mass M (in kg mol⁻¹) with R. At the same temperature, lighter molecules move faster: they need a larger ⟨c²⟩ to have the same kinetic energy. For one gas, c_(r.m.s.) ∝ square root of T, so four times the thermodynamic temperature means twice the r.m.s. speed.

R.m.s. speed for nitrogen and helium

Both square-root curves increase from 100 to 700 kelvin. At 300 kelvin nitrogen is about 517 metres per second and helium about 1367 metres per second.

Scroll across the figure to read all labels.

Both square-root curves increase from 100 to 700 kelvin. At 300 kelvin nitrogen is about 517 metres per second and helium about 1367 metres per second.Both square-root curves increase from 100 to 700 kelvin. At 300 kelvin nitrogen is about 517 metres per second and helium about 1367 metres per second.
Calculated ideal-gas predictions cᵣₘₛ = square root of (3RT/M) using R = 8.31 J mol⁻¹ K⁻¹. Nitrogen: M = 0.0280; helium: M = 0.00400 kg mol⁻¹. These are model values, not measurements.
Open full-size graph
What to noticeAt a common temperature, helium is square root of 7 times faster in r.m.s. speed, although both gases have the same mean translational energy.
View figure data
Figure data: R.m.s. speed for nitrogen and helium
Temperature, T (K)Nitrogen, M = 0.0280Helium, M = 0.00400
100298.4789.5
105305.8809.0
110313.0828.0
115320.0846.6
120326.9864.8
125333.6882.6
130340.2900.1
135346.7917.3
140353.1934.1
145359.3950.6
150365.4966.9
155371.5982.9
160377.4998.6
165383.31014.1
170389.11029.3
175394.71044.4
180400.31059.2
185405.91073.8
190411.31088.2
195416.71102.4
200422.01116.5
205427.21130.3
210432.41144.0
215437.51157.6
220442.61171.0
225447.61184.2
230452.51197.3
235457.41210.2
240462.31223.0
245467.11235.7
250471.81248.2
255476.51260.7
260481.11273.0
265485.71285.2
270490.31297.2
275494.81309.2
280499.31321.0
285503.71332.8
290508.11344.4
295512.51355.9
300516.81367.4
305521.11378.7
310525.41390.0
315529.61401.2
320533.81412.2
325537.91423.2
330542.01434.1
335546.11445.0
340550.21455.7
345554.21466.4
350558.21476.9
355562.21487.5
360566.21497.9
365570.11508.3
370574.01518.6
375577.81528.8
380581.71538.9
385585.51549.0
390589.31559.1
395593.01569.0
400596.81578.9
405600.51588.8
410604.21598.5
415607.91608.3
420611.51617.9
425615.11627.5
430618.81637.1
435622.31646.6
440625.91656.0
445629.51665.4
450633.01674.7
455636.51684.0
460640.01693.2
465643.41702.4
470646.91711.5
475650.31720.6
480653.71729.6
485657.11738.6
490660.51747.5
495663.91756.4
500667.21765.3
505670.51774.1
510673.91782.9
515677.21791.6
520680.41800.2
525683.71808.9
530686.91817.5
535690.21826.0
540693.41834.5
545696.61843.0
550699.81851.5
555703.01859.8
560706.11868.2
565709.31876.5
570712.41884.8
575715.51893.1
580718.61901.3
585721.71909.5
590724.81917.6
595727.81925.7
600730.91933.8
605733.91941.8
610737.01949.8
615740.01957.8
620743.01965.7
625746.01973.7
630748.91981.5
635751.91989.4
640754.91997.2
645757.82005.0
650760.72012.7
655763.72020.5
660766.62028.2
665769.52035.8
670772.42043.5
675775.22051.1
680778.12058.7
685781.02066.2
690783.82073.7
695786.62081.2
700789.52088.7

Worked example 1

Nitrogen at room temperature

Problem

Nitrogen, of molar mass 0.0280 kg mol⁻¹, is at 300 K. Find the mean translational kinetic energy of a molecule and the r.m.s. speed.

Show full solution

  1. Mean kinetic energy

    Method

    Use (1/2)m⟨c²⟩ = (3/2)kT.

    Reason

    The mean energy per molecule depends only on the temperature.

    Working

    Answer

    (3/2)kT = (3/2)(1.38 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J

  2. R.m.s. speed

    Working

    Answer

    The molar mass is given, so use R: c_(r.m.s.) = square root of (3RT/M) = square root of (3(8.31)(300)/0.0280) = 517 m s⁻¹

Check your understanding 1 Find the r.m.s. speed

Three molecules have speeds 100, 200 and 300 m s⁻¹. Calculate their r.m.s. speed to three significant figures.

Show the steps Optional working
Unit: m² s⁻²
Unit: m s⁻¹
Show answer

c_(r.m.s.) = square root of ((100² + 200² + 300²)/3) = 216 m s⁻¹. The mean speed is 200 m s⁻¹; squaring that mean would discard the spread in speeds.

Guided practice 2

Heating from 20 °C to 40 °C

About 3 min

Problem

A gas is heated from 20.0 °C to 40.0 °C. By what factor does the mean kinetic energy of its molecules increase? By what factor does the r.m.s. speed increase? Give both factors to three significant figures.

Find both factors

Show the steps Optional working

Hints

Hint 1

kelvin

Which temperature scale belongs in the energy and speed equations?

Show solution step by step

  1. Temperature ratio

    Reason

    The relations use thermodynamic temperature.

    Working

    Answer

    T₂/T₁ = 313.15/293.15 = 1.068

  2. Factors

    Working

    Answer

    The mean kinetic energy rises by a factor of 1.07, and the r.m.s. speed by square root of (313.15/293.15) = 1.03, both to three significant figures. Neither doubles: 40 °C is not twice as hot as 20 °C.

Check your understanding 2 Twice the speed

The r.m.s. speed of the molecules of a gas doubles. By what factor have the thermodynamic temperature and the mean kinetic energy changed? Does a faster gas always have the higher temperature?

Does the faster of two different gases always have higher temperature?
Show answer

Both increase four times, because they are proportional to ⟨c²⟩. Not when comparing different gases: temperature depends on m⟨c²⟩, so a gas of heavy, slow molecules can be hotter than a gas of light, fast ones.

Spot the mistake 3

Squaring the mean speed

About 4 min

Problem

In a group of identical molecules, each of mass 4.00 × 10⁻²⁶ kg, half move at 200 m s⁻¹ and half at 600 m s⁻¹. A learner finds the mean speed, 400 m s⁻¹, and gives the mean kinetic energy as (1/2)m(400)² = 3.20 × 10⁻²¹ J. Find the correct mean kinetic energy.

Find the mean kinetic energy

Show the steps Optional working
Unit: m² s⁻²
Unit: J
Show solution step by step

  1. Mean of the squares

    Reason

    Each molecule’s kinetic energy depends on its own speed squared.

    Working

    Answer

    ⟨c²⟩ = (200² + 600²)/2 = (40 000 + 360 000)/2 = 2.00 × 10⁵ m² s⁻²

  2. Mean kinetic energy

    Working

    Answer

    (1/2)m⟨c²⟩ = (1/2)(4.00 × 10⁻²⁶)(2.00 × 10⁵) = 4.00 × 10⁻²¹ J, and c_(r.m.s.) = square root of (2.00 × 10⁵) = 447 m s⁻¹

Try it yourself 4

Can the slower gas be hotter?

Minimal support

Problem

Helium atoms (M = 0.00400 kg mol⁻¹) in one container have an r.m.s. speed of 900 m s⁻¹. Nitrogen molecules (M = 0.0280 kg mol⁻¹) in another have an r.m.s. speed of 500 m s⁻¹. Find the temperature of each gas.

Find both temperatures

Unit: K
Unit: K

Hints

Hint 1

rearrange

Rearrange the r.m.s. speed equation to make temperature the subject.

Show solution step by step

  1. Temperatures

    Reason

    Each gas has its own molar mass and speed.

    Working

    Answer

    Helium: T = 0.00400(900)²/3(8.31) = 130 K. Nitrogen: T = 0.0280(500)²/3(8.31) = 281 K.

  2. Which is hotter

    Working

    Answer

    The nitrogen is hotter, and its molecules have about 2.2 times the mean kinetic energy, even though they move more slowly. Speed alone does not decide temperature; mass matters too.

Exam-style question 1

Temperature from molecular energy

5 marks

Examination question

(a) Use pV = (1/3)Nm⟨c²⟩ and pV = NkT to show that the mean translational kinetic energy of a molecule of an ideal gas is (3/2)kT. [2 marks]

(b) The molecules of a gas have a mean translational kinetic energy of 1.242 × 10⁻²⁰ J. Find the temperature of the gas. State whether molecules of twice the mass, at the same temperature, would have more kinetic energy. [3 marks]

Write your answer before viewing the mark scheme

Show solution step by step

  1. (a) Derivation

    2 marks

    Method

    Equate the two expressions for pV.

    Reason

    They describe the same gas, so N cancels.

    Working

    Answer

    (1/3)Nm⟨c²⟩ = NkT, so (1/3)m⟨c²⟩ = kT. Multiplying by 3/2: (1/2)m⟨c²⟩ = (3/2)kT.

  2. (b) Temperature and mass

    3 marks

    Working

    Answer

    T = (2(1.242 × 10⁻²⁰))/(3(1.38 × 10⁻²³)) = 600 K. No: at the same temperature the mean kinetic energy is the same, (3/2)kT; the heavier molecules have half the mean-square speed.

Common mistakes

  • Squaring the mean speed instead of finding the mean of the squared speeds.
  • Thinking heavier molecules have more kinetic energy at the same temperature.
  • Thinking every molecule has the energy (3/2)kT; that is the mean.
  • Using a molar mass with k, or the mass of one molecule with R.
  • Using Celsius temperatures in (3/2)kT.

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