Formulation of Kinetic Theory of Gas

Formulation of kinetic theory of gas
Kinetic gas equation
    After knowing the experimental gas lows, it is of interest to develop a theoretical model based on the structure and properties of gases, which can be correlated to the experimental facts.
    Fortunately, such theory has been developed for the formulation of the kinetic theory of gases based upon the certain postulates which are supposed to be applicable to an Ideal gas.

Assumptions of the kinetic theory of gases

  1. The gas is composed of very small discrete particles, now called molecules. For gas, the mass and size of the molecules are the same and different for different gases.
  2. The molecules are moving in all directions with the verity of speeds. Some are very fast while others are slow.
  3. Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (intermolecular collision). These collisions are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
  4. The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
  5. There exist no intermolecular attraction especially at low pressure, that is one molecule that can exert pressure independent of the influence of other molecules.
  6. The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas.
  7. This explains Boyle's law since when the volume is reduced, wall collision becomes more frequent and pressure is increased.
  8. Through the molecular velocity is constantly changing due to the intermolecular collision, the average kinetic energy(є) of the molecules remains fixed at a given temperature. This explains the Charl's law that when T is increased, velocity is increased, wall collision becomes more frequent and pressure(P) is increased when T kept constant or Volume(V) is increased when P kept constant.

Root mean square velocity

    Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds.
CRMS² = (N₁C₁² + N₂C₂² + ...)/N

Formulation of Kinetic gas equation

    Let us take a cube of edge length l containing N molecules of gas of molecular mass m and RMS speed is CRMS at temperature T and pressure P.
Formulation of kinetic theory of gas
Kinetic theory of gas

    Let in gas molecules, N₁ have velocity C₁, N₂ have velocity C₂, N₃ have velocity C₃, and so on.
    Let us concentrate our discussion to a single molecule among N₁ that has resultant velocity C₁ and the component velocities are Cx, Cy, and Cz.
C₁² = Cx² + Cy² + Cz²
    The molecule will collide walls A and B with the component velocity Cx and other opposite faces by Cy and Cz.
Change of momentum along X-direction for a single collision,
= m Cx - (- m Cx) = 2 m Cx

Rate of change of momentum of the above type of collision,
= 2 mCx × (Cx/l)

= 2 mCx²/l
    Similarly, along Y and Z directions, the rate of change of momentum of the molecule is 2 mCy²/l and 2 mCz²/l respectively.
Total rate change of momentum for the molecule,
= 2 mCx²/l + 2 mCx²/l +2 mCz²/l
= 2 (m/l) (Cx² + Cy² + Cz²)

= 2 mC₁²/l

For similar N₁ molecules, it is 2 mN₁C₁²/l

Taking all the molecules of the gas, the total rate of change of momentum,
= (2 mN₁C₁²/l)+(2 mN₂C₂²/l)+(2 mN₃C₃²/l)+ ..
= 2 mN {(N₁C₁² + N₂C₂² + N₃C₃² ..)/N}

= 2 mN CRMS²

where CRMS² = root means square velocity of the gases.
    According to Newton's 2nd low of motion, the rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6l² = 2 mN CRMS²/l
or, P × l³ = 1/3 m N CRMS²

∴ PV = 1/3 m N CRMS²

Here, l³ = volume of the cube contains gas molecules.

The other form of the equation is,
P = 1/3 × (mN/V) × CRMS²

∴ P = 1/3 d CRMS²
where (mN/V) is the density(d) of the gas molecules.
This equation is also valid for any shape of the gas container.

Root mean square velocity

Let us apply the kinetic equation for a 1-mole ideal gas.
In that case mN = mN₀ = M
Ideal gas equation,
PV = RT.
Hence from the kinetic gas equation, PV = 1/3 m N CRMS²
or, RT = 1/3 m N CRMS²

or, CRMS² = 3RT/M

∴ CRMS = √3RT/M
    Thus root means square velocity depends on the molar mass(M) and temperature(T) of the gas.

Average kinetic energy

The average kinetic energy(Ē) is defined as,
Ē = 1/₂ m CRMS²

Again from the kinetic gas equation,
PV = 1/3 m N CRMS²
or, PV = ²/3 N × 1/3 m CRMS²
or, PV = ²/3 N Ē

For 1 mole ideal gas,PV = RT and N = N₀
Thus RT = ²/3 N₀ Ē

or, Ē = ³/₂R/N₀T
∴ Ē = ³/₂ kT
Where k = R/N0 and is known as the Boltzmann constant.
Its value is 1.38 × 10-23 JK⁻¹

The total kinetic energy for 1 mole of the gas is,

ETotal = N₀ (Ē) = ³/₂RT
    Thus average kinetic energy is dependent on T only and it is not dependent on the nature of the gas.

Problems solutions

Problem
    Calculate the pressure exerted by 10²³ gas particles each of the mass 10⁻²² gm in a container of volume 1 dm³. The root means square speed is 10⁵ cm sec⁻¹.
Solution
    We have, N = 10²³, m = 10⁻²² gm = 10⁻²⁵ Kg, V = 1 dm³ = 10⁻³ m³ and CRMS² = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.
Therefore, from the kinetic gas equation,
PV = 1/3 mN CRMS²
or, P = 1/3 × (m N/V) × CRMS²
Putting the value we have,
P = (1/3)(10⁻²⁵ Kg×10²³/10⁻³ m³) × (10³ m sec⁻¹)²
∴ P = 0.333 × 10⁷ atm
Problem
    Calculate the root mean square speed of oxygen gas at 27⁰ C.
Solution
We know that,
CRMS² = (3RT/M)

Here, M = 32 gm mol⁻¹, and T = 270 C = (273+27)K = 300 K.

Thus,CRMS² = (3 × 8.314 × 107 erg mol⁻¹ K⁻¹ × 300 K)/(32 gm mol⁻¹)

∴ CRMS = 48356 cm sec⁻¹
Problem
    Calculate the RMS speed of NH3 at N.T.P.
Answer
    At N.T.P, V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹, P = 1 atm = 101325 Pa and M = 17 × 10⁻³ Kg mol⁻¹.
Thus, CRMS² = 3RT/M
= (3 × 101325 × 22.4 × 10⁻³) /(17 × 10⁻³)
∴ CRMS = 632 m sec⁻¹
Problem
    How the root means square velocity for oxygen compares with that of the hydrogen?
Answer
We know that, CRMS² = 3RT/M
Hence at a given temperature,
(CRMS of H₂)²/(CRMS of O₂)² = MO₂/MH₂
= 32/2 = 16
That is CRMS of O₂ = 4 × CRMS of H₂.
Problem
    Calculate the kinetic energy of translation of 8.5 gm NH₃ at 27⁰ C.
Solution
We know that total kinetic energy for 1 mole of the gas is,
ETotal = (3/2)RT

= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K)
= 900 cal mol⁻¹
Again 8.5 gm NH3 = (8.5/17) mol = 0.5 mol
Thus the kinetic energy of 8.5 gm NH3 at 27⁰ C is, (0.5 × 900) cal

= 450 cal
Problem
    Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 K.cal mol⁻¹. At what temperature the molecules have this value of KE?
Solution
T = 673.9K or 400.9⁰ C

Assumptions and formulation of kinetic theory of gas, root mean square velocity and average kinetic energy with problems solutions

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