Kinetic Interpretation of Temperature and RMS Speed of Gas Molecules

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In Kinetic theory, the kinetic interpretation of temperature states that the average kinetic energy of the gas molecules of an ideal gas is directly proportional to the absolute temperature of the molecules.

  • The average kinetic energy is independent of the pressure, volume, and nature of the gas.
  • The energy (per molecule) for any monatomic atoms like argon and diatomic molecules like chlorine is equal to (3/2) kBT.
  • The kinetic interpretation of temperature relates to the internal energy and temperature of a molecule.
  • It is determined that the average velocity of the molecules increases with the increase in the temperature of the gas.
  • The kinetic molecular theory will explain the Boyle law, Charles law, Avogadro law and Dalton law.

Key Terms: Kinetic Interpretation of Temperature, RMS Speed of Gas Molecules, Average Speed of The Molecules of Gas, Kinetic energy, Internal energy, Ideal gas equation, Boltzmann constant, Gas Constant.


Kinetic Theory of an Ideal Gas

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The kinetic theory of an ideal gas is based on the molecular picture of matter. According to which

  • A given amount of gas is a mixture of a very large number of identical molecules of the order of Avogadro’s number.
  • The molecules move randomly in all directions.
  • At ordinary temperature and pressure, the size of the molecules is very small compared to the distance between them.
  • Thus, the interaction between them is negligible. Hence they move according to Newton’s law of motion.
  • Except during collisions, there is no force exerted between the molecules.
  • The collisions of the molecules against each other or with the walls of the container are perfectly elastic.
  • Therefore, though their velocities change, momentum and the kinetic energy of the molecules are conserved during the collision.

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Kinetic Interpretation of Temperature

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Kinetic Interpretation of Temperature is used to express relation between the temperature and the kinetic energy of the molecules. For gas molecules, as the temperature increases, the kinetic energy of the molecules also increases.

  • The kinetic energy increases as molecules get more energy for movement.
  • The results is similar in case of liquids.
  • In the case of solids, with an increase in temperature, the molecules start vibrating with greater speed.

Derivation of Kinetic Interpretation of Temperature

Consider 1 mole of an ideal gas enclosed in a cube of volume V and the molecules of the gas are moving with velocity v, then the pressure exerted by the ideal gas on the wall of the container is given by

\(P=\frac{1}{3}mnv^2 \)

Where

  • v = RMS (root mean square) velocity 
  • P = pressure 
  • m = mass of one molecule 
  • n = NA/V = number of molecules (NA) per unit volume (V)
  • NA = Avagadro's number = number of molecules in one mole of gas

\(\Rightarrow P=\frac{1}{3}m\frac{N_A}{V}v^2 \)

\(\Rightarrow PV=\frac{1}{3}mN_Av^2\)   ...(i)

From the ideal gas equation, we have

PV = μRT 

Where

  • μ = number of moles of the gas
  • R = Gas constant
  • T = Absolute temperature

For one mole, μ = 1

⇒ PV = RT

Substituting the above value in equation (i), we get

\( RT=\frac{1}{3}mN_Av^2\)

\(\)\(\Rightarrow \frac{RT}{N_A}=\frac{1}{3}mv^2\)

\(\Rightarrow \frac{RT}{N_A}=\frac{2}{3} \times \frac{1}{2} mv^2\)

\(\Rightarrow \frac {1}{2}mv^2= \frac {3}{2}\frac {RT}{N_A}\)    ...(ii)

Now the average kinetic energy of the molecules of the gas is given by

Kinetic energy,  

\((KE)_{avg}=\frac{1}{2}mv^2\)

Using equation (ii), we get

The average kinetic energy,

 \((KE)_{avg}= \frac {3}{2}\frac {RT}{N_A}\)

But, R = NAKB

Where

  • KB is the Boltzmann constant 
  • NA is Avagadro's number

Therefore, the average kinetic energy of the molecules of the gas is given by

\((KE)_{avg}= \frac {3}{2}\frac {N_AK_B}{N_A}T\) 

\(\Rightarrow (KE)_{avg}= \frac {3}{2}K_BT\)

The above equation shows that the average kinetic energy of a molecule of gas is directly proportional to the absolute temperature of the gas. This is the Kinetic interpretation of temperature. i.e.

\((KE)_{avg}=\frac {1}{2}mv^2= \frac {3}{2}K_BT\)

OR

\((KE)_{avg}\propto T\)

Kinetic explanation of Boyle's law

Boyle's law can be explained using kinetic molecular theory. According to the law, pressure depends upon the number of times the molecules strike the surface of the container.

  • When gas molecules are compressed to smaller volume, then in such case, the number of molecules are found at a smaller surface area.
  • This will result in an increase in the number of strikes per unit of area and pressure of the area.

Kinetic explanation of Charles' law

According to kinetic molecular theory, the average kinetic energy of molecules increases with an increase in temperature. As molecules move rapidly with pressure being constant, they must stay farther apart in such a situation.

  • This will result in a collision of molecules with the surface of the container.
  • It is later compensated by an increase in the area of the surface as the gas expands.

Kinetic explanation of Avogadro's law

According to kinetic molecular theory, the number of molecules colliding with the walls per unit of time increases with respect to the increase in the number of gas molecules in a closed container.

  • The volume increases with pressure remains constant.
  • As a result of which, molecules will hit the walls less frequently.

Kinetic explanation of Dalton's law

According to Dalton law of partial pressure every gas molecules is independent of other gas molecules. This can be explained with kinetic molecular theory where gas molecules have zero volume. According to the percentage of molecules it represents, each molecule adds a certain amount of pressure to the overall pressure inside the container.


Root Mean Square Speed (RMS Speed)

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Root mean square speed or RMS speed refers to the square root of the mean of the squares of the random speeds of the individual molecules of the gas. The speed will express the velocity of the gas molecules in terms of their temperature and molar mass.

  • It is given by

\(v_{rms}=\sqrt{v^2}\)   ...(i)

  • From the kinetic interpretation of temperature, the average kinetic energy of a molecule of gas is given by

\((KE)_{avg}=\frac {1}{2}mv^2= \frac {3}{2}K_BT\)

\(\Rightarrow v^2= \frac{3K_BT}{m}\)   ...(ii)

\(\Rightarrow v^2 \propto T\)

  • Hence, using equation (i), we get

\(v_{rms} \propto \sqrt{T}\)

  • i.e. rms velocity of the gas molecules is directly proportional to the square root of absolute temperature.
  • Also, using equation (ii), we get

\( v_{rms}= \sqrt{\frac{3K_BT}{m}}\)

  • But, KB = R\NA

\( \Rightarrow v_{rms}= \sqrt{\frac{3RT}{mN_A}}\)

Substituting mNA = M, the molar mass of the gas, we get

\(v_{rms}= \sqrt{\frac{3RT}{M}}\)


Average Speed of the Molecules of Gas

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The average speed of the molecules of a gas is the arithmetic mean of the speed of the molecules of a given gas at a given temperature. Let a container contains one mole of a gas and v1, v2, v3, ……… be the velocity of each molecule of the gas.

  • Then the average speed of a molecule of gas is given by

\(v_{av}=\frac {|v_1|+|v_2|+|v_3|+.....}{N_A}=\sqrt{\frac {8RT}{\pi M}}=\sqrt{\frac {8K_BT}{\pi m}}\)

Where

  • NA = Avagadro’s number
  • M = mass of one mole of gas
  • m = mass of one molecule of gas

Most Probable Speed of The Molecules of Gas

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The most probable speed is the speed possessed by the maximum number of molecules in a given gas at a given temperature. It is given by

\(v_{mp}=\sqrt{\frac {2RT}{M}}=\sqrt{\frac {2K_BT}{m}}=\sqrt{\frac {2}{3}}v_{rms}\)

Where

  • M = mass of one mole of gas
  • m = mass of one molecule of gas

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Things to Remember

  • Kinetic interpretation of temperature is based on the assumption that the gas is an ideal gas. 
  • The rms speed refers to the root mean square speed and it is calculated as the root of the average of the square of velocities.
  • Root mean square speed is used to calculate the speed of particles that are in a constant state of motion.
  • The ideal gas equation is PV = nRT.
  • It is assumed that in the case of the Kinetic Theory of Gases, then, no force will be acting on the molecules except during elastic collisions.
  • Temperature is directly proportional to the square of rms velocity; T ∝ v2

Sample Questions

Ques: Calculate the RMS speed of O2 molecule at 20°C, where R = 8.315 J/mol K? (3 marks)

Ans: Given – 

T = 273 + 20 = 293 K

M = Oxygen O2 = 16 × 2 = 32 g/mole = 32 × 10-3 kg/mole

O2 – vrms = v = \(\sqrt{\frac{3RT}{M}}\) = \(\sqrt{\frac{3×8.315×293}{32× 10^{-3}}}\)

O2 – vrms = 478 m/s

Ques: Calculate the RMS speed of CO2 molecule at 20°C, where R = 8.315 J/mole K? (3 marks)

Ans: Given – 

T = 273 + 20 = 293 K

M = Carbon dioxide CO2 = 12 + (16 × 2) = 44 g/mole = 44 × 10-3 kg/mole

CO2 – vrms = v = \(\sqrt{\frac{3RT}{M}}\) = \(\sqrt{\frac{3×8.315×293}{44× 10^{-3}}}\)

CO2 – vrms = 408 m/s

Ques: Calculate the RMS speed of H2 molecule at 20?C, where R = 8.315 J/mole K? (3 marks)

Ans: Given – 

T = 273 + 20 = 293K

M = Hydrogen H2 = 1 × 2 = 2 g/mole = 2 × 10-3 kg/mole

H2 – vrms = v = \(\sqrt{\frac{3RT}{M}}\) = \(\sqrt{\frac{3×8.315×293}{2× 10^{-3}}}\)

H2 – vrms = 1912 m/s

Ques: How to determine velocity of gas molecule OR To prove vrms = \(\sqrt{\frac{3RT}{M}}\) ?(5 marks)

Ans: We have two equations, 

P = \(\frac{1}{3}\frac{mn}{V}v^2\)

PV = \(\frac{1}{3}mnv^2\)

Ideal gas equation, PV = nRT

Equate the equations

\(\frac{1}{3}mnv^2\) = nRT

v2 = \(\frac{3RT}{M}\)

vrms = v = \(\sqrt{\frac{3RT}{M}}\)

K = \(\frac{R}{N_A} \)

NA – Avogadro number

R = K × NA

v = \(\sqrt{\frac{3\ K\times N_A \times T}{M}}\)

v = \(\sqrt{\frac{3\ K\ T}{M/N_A}}\)

\(\frac{M}{N_A}\) = m

vrms = v = \(\sqrt{\frac{3RT}{m}}\)

Hence proved.

Ques: How the temperature is related to root mean square (rms) velocity OR to prove T v2 OR derive the kinetic interpretation of temperature? (5 marks)

Ans: P = \(\frac{1}{3}\frac{mn}{V}v^2\)....................................................... (1)

v – RMS (root mean square) velocity

P – pressure

V – volume

m – mass of one molecule or M – molar mass of molecule

n – number of moles

In this equation, suppose we take one mole of molecule

Let gas be one mole and the number of molecules N, 

PV= \(\frac{1}{3}mNv^2\) ................................................... (2)

By the ideal gas equation, PV = nRT 

For one mole, n=1

PV = RT

RT = \(\frac{1}{3}mNv^2\)

\(\frac{RT}{N}\) = \(\frac{1}{3}mv^2\) .....................................................(3)

We are going to determine kinetic energy of one molecule

We know that, 

Kinetic energy (KE) = \(\frac{1}{2}mv^2\)

m – mass and v – velocity

Now rewrite the equation (3) by multiply and divided by 2

\(\frac{RT}{N}\) = \(\frac{2}{3} \times \frac{1}{2} \times mv^2\)

\(\frac{3}{2} \frac{R}{N} T\) = Kinetic energy of one molecule

\(\frac{R}{N}\) = constant = KB – Boltzmann constant 

While, R is the gas constant and N is the number of molecules, so constant divided by constant is constant

\(\frac{3}{2} K_B \times T\) = KE of one molecule

i.e.

KE of one atom = \(\frac{3}{2} KT\)

\(\frac{3}{2} KT\) = constant

So, energy solely depends upon the temperature of the body

\(\frac{1}{2}mv^2\) = \(\frac{3}{2} KT\)

m – mass of one molecule i.e. constant

Therefore,

T ∝ v2 ................................................................. (4)

From equation (4), 

Temperature is directly proportional to square of rms velocity; this is the temperature and velocity related to each other. 

Ques: Define kinetic theory of gas molecule? (2 marks)

Ans: The kinetic theory states that the average kinetic energy of the gas molecules of an ideal gas is directly proportional to the absolute temperature of the molecules. Kinetic theory is independent of pressure, volume and nature of gas. This kind of interpretation is called kinetic interpretation of temperature. 

Ques: To find the pressure equation of kinetic theory of an ideal gas OR Derive the pressure equation of gas? (5 marks)

Ans:

Kinetic Model of an Ideal Gas 1 Kinetic Model of an Ideal Gas 2

Kinetic Model of an Ideal Gas

Gas molecule in x, y and z direction in box with length L

The gas molecule of mass m, having three velocity component in x,y,and z direction i.e. vx , vy and vz

The average of the velocities is,

v2 (rms) = vx+ vy+ vz2 .............................. (A)

Now we going to calculate pressure, 

First we calculate momentum (p) of gas molecule in x direction,

px = m vx , ..................................................... (B)

now change in momentum, 

Δpx = 2 m vx ................................................. (C)

Impulse, I = Δ p = F Δt

t – time , F – force, Δ p – change in momentum

F = \(\frac{\Delta p}{\Delta t}\)

Now to define Δt,

Distance (d) = velocity (v) × time (t)

time (t) = \(\frac{Distance (d)}{velocity (v)}\) = \(\frac{2L}{vx}\)

F = \(\frac{\Delta p}{\Delta t}\) = \(\frac{2 m vx} {\Delta t}\)

F = \(\cfrac{2 m vx} {\frac{2L}{vx}}\) ............................................... (D)

After simplified equation (D), we get

F = \(\frac{mvx^2}{L}\) .................................................. (E)

As, vx = vy = vz

From (A), 

v2 (rms) = vx+ vy+ vz2 = 3vx2

vx= \(\frac{1}{3}\) v2 ................................................. (F)

From equation (E) and (F),

F = \(\frac{1}{3}\frac{m}{L}v^2\)

If we calculate the force of ‘n’ molecule of the gas 

FN = \(\frac{mv^2N}{3L}\)

Pressure (P) = \(\frac{Force (F)}{Area (A)}\)

Area = L2

PN = \(\frac{F_N}{A}\)

P = \(\frac{1}{3}\frac{mn}{V}v^2\) ............................................ (V= L3)

Hence proved.

Ques: Calculate the RMS speed of NO2 molecule at 20°C, where R = 8.315 J/mole K? (3 marks)

Ans: Given – 

T = 273 + 20 = 293 K

M = Nitrogen Oxygen NO2 = 14.0067 + (16 × 2) = 46.0067 g/mole = 46.0067 × 10-3 kg/mole

NO2 – vrms = v = \(\sqrt{\frac{3RT}{M}}\) = \(\sqrt{\frac{3×8.315×293}{46.0067× 10^{-3}}}\)

NO2 – vrms = 398.57 m/s

Ques: Calculate the RMS speed of N2 molecule at 20 degree C, where R = 8.315 J/mole K? (3 marks)

Ans: Given – 

T = 273 + 20 = 293K

M = N2 = 16 × 2 = 32 g/mole = 32 × 10-3 kg/mole

N2 – vrms = v = \(\sqrt{\frac{3RT}{M}}\) = \(\sqrt{\frac{3×8.315×293}{14.0067× 10^{-3}}}\)

N2 – vrms = 722.36 m/s

Ques: A boy is transporting a cupboard of mass 9kg with a kinetic energy of 180 Joules. Find the velocity at which the boy is running? (2 marks)

Ans: Mass, m = 9 Kg

Kinetic energy K.E = 180 J

The man is running with a velocity of 6.32 m/s

Ques: How kinetic energy is used to determine temperature? (2 marks)

Ans: Kinetic energy and temperature are connected. The average kinetic energy of the particles within an object is measured by its temperature. The average particle speed of anything increases with its temperature.


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CBSE CLASS XII Related Questions

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      • 2.
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              • 4.
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                  • 5.
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                    • 6.
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                        CBSE CLASS XII Previous Year Papers

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