### Particles in a solid-liquid and gas

The common behavior of different gases suggested that the internal structure in all gases must be similar. Study in the seventeenth-century explained Gassendhi and Hooke explained the physical phenomena of gases on the assumption of the existence of rapidly moving independent minute particles. Bernoulli in 1738 the first time explained the properties of gases on a mechanical basis.In solid the molecules or particles are held very closely together and are entirely devoid of any translatory motion. If heat is supplied to solid, it takes the form of vibrational motion with the rise of temperature. With the further increases the thermal energy, the vibrational motion rises to such extent the molecules break down to transform into a liquid.

Further absorption of heat causes the particles to brack away from the restraining forces holding together and the particles move away from the liquid to the gaseous state. The gases then are essentially composed freely moving particles.

These basic ideas were at the root of the theory to explain the behavior of gases called the kinetic theory of gases. In the nineteenth century, the effort of Joule, Kronig, Clausius, Boltzmann, and Maxwell, the theory succeeded in attaining a rigid mathematical form.

Fortunately, such theory has been developed for the formulation of the kinetic gas equation based upon certain postulates which are supposed to be applicable to an ideal gas.

#### Postulates of kinetic theory for gaseous molecules

- Gas molecules are composed of very small discrete particles. In any one gas, all the molecules are of the same size and mass, but these differ gas to gas.
- Gas molecules within the container are moving in all directions with a verity of speeds. Some are very fast while others are slow.
- Due to random motion, the gas molecules are executing collision with the walls of the container called wall collision and with themselves called intermolecular collision.

These collisions are perfectly elastic and so there occurs no loss of**kinetic energy**or momentum of the molecules by this collision. - Gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
- There exist no intermolecular attraction especially at low pressure, that is one molecule that can exert pressure independent of the influence of other molecules.
- 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.
- This explains Boyle's law since when the volume is reduced at a constant temperature, wall collision becomes more frequent and pressure is increased.
- Through the molecular velocity constantly changing due to the intermolecular collision, the average kinetic energy of the gas molecules remains fixed at a given temperature.

#### Charles law from kinetic theory

When temperature raised, the molecules would move more vigorously resulting in a larger number of impacts on the wall of the container at constant volume. That is why we find an increase of pressure with a rise in temperature at constant volume. This is Charles's law.#### What is RMS for gas molecules?

RMS or root mean square speed is defined as the square root of the average of the squares of speeds.C

_{RMS}² = (N₁C₁² + N₂C₂² + ...)/N### Formulate kinetic gas equation for the ideal gases

Let us take a cubic gas container with edge length l containing N molecules of gas of mass m and RMS speed C_{RMS}at temperature T and pressure P.

The molecules are moving constantly with different velocities in different directions bombarding on the walls of the cube.

Gas molecules, N₁ have velocity C₁, N₂ have velocity C₂, N₃ have velocity C₃, and so on.

Concentrate our discussion on a single molecule among N₁ that has resultant velocity C₁ and the component velocities are Cx, Cy, Cz.

Component velocity of gas molecules |

∴ C₁² = C

_{x}² + C_{y}² + C_{z}²The molecule will collide walls A and B of the container 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 for this collision,

= 2 mCx × (Cx/l)

= 2 mCx²/l.

Similarly, along Y and Z directions, the rate of change of momentum for the molecule 2 mCy²/l and 2 mCz²/l respectively.= m Cx - (- m Cx) = 2 m Cx.

Rate of change of momentum for this collision,

= 2 mCx × (Cx/l)

= 2 mCx²/l.

Hence the net momentum imparted on all six walls of the container by the collision of the gas molecules,

= 2 mCx²/l + 2 mCx²/l +2 mCz²/l

= 2 (m/l) (Cx² + Cy² + Cz²)

= 2 mC₁²/l

For N₁ molecules, change of momentum

= 2 mN₁C₁²/l.

= 2 (m/l) (Cx² + Cy² + Cz²)

= 2 mC₁²/l

For N₁ molecules, change of momentum

= 2 mN₁C₁²/l.

Consider all the molecules of the gas present in the cubic container, the total rate of change of momentum,

= (2 mN₁C₁²/l)+(2 mN₂C₂²/l)+(2 mN₃C₃²/l)+ ..

= 2 mN {(N₁C₁² + N₂C₂² + N₃C₃² ..)/N}

= 2 mN C

where C

= 2 mN {(N₁C₁² + N₂C₂² + N₃C₃² ..)/N}

= 2 mN C

_{RMS}²where C

_{RMS}= root means square velocity of the gases.According to Newton's second law of motion, the rate of change of momentum due to wall collision is equal to force developed within the gas molecules.

P × 6l² = 2 mN C

or, P × l³ =

∴ PV =

where l³ = volume of the cubic container.

_{RMS}²/lor, P × l³ =

^{1}/_{3}m N C_{RMS}²∴ PV =

^{1}/_{3}m N C_{RMS}²where l³ = volume of the cubic container.

Kinetic gas equation |

#### Another form of the equation

P =

∴ P =

where (mN/V) is the density(d) of the gas molecules.

^{1}/_{3}× (mN/V) × C_{RMS}²∴ P =

^{1}/_{3}d C_{RMS}²where (mN/V) is the density(d) of the gas molecules.

This equation is valid for any shape of the gas container.

Problem

Calculate the pressure exerted by 10²³ gas molecules each of the mass 10⁻²² gm in a container of volume 1 dm³. Given RMS speed of the gas molecule 10⁵ cm sec⁻¹.

Solution

Numer of gas molecules (N) = 10²³

Mass of the gas molecule (m) = 10⁻²² gm

= 10⁻²⁵ Kg

Volume (V) = 1 dm³ = 10⁻³ m³

CRMS = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.

Kinetic gas equation,

PV = 1/3 mN C

or, P = (m × N × C

∴ P = (10⁻²⁵ × 10²³ × 10⁻⁶)/(3 × 10³)

∴ P = 0.333 × 10⁷ Pascal.

Mass of the gas molecule (m) = 10⁻²² gm

= 10⁻²⁵ Kg

Volume (V) = 1 dm³ = 10⁻³ m³

CRMS = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.

Kinetic gas equation,

PV = 1/3 mN C

_{RMS}²or, P = (m × N × C

_{RMS}²)/(3 × V)∴ P = (10⁻²⁵ × 10²³ × 10⁻⁶)/(3 × 10³)

∴ P = 0.333 × 10⁷ Pascal.

#### Velocity formula for gas molecules

The kinetic relations and ideal gas law may be used to formulate the velocity of the gas moleculesKinetic relation for a 1-mole ideal gas

PV =

where mN = mN₀ = M.

Ideal gas law for one-mole gas

PV = RT.

∴ PV =

or, RT =

or, C

∴ C

Thus root means square velocity depends on the molecular weight and temperature of the gas molecules.PV =

^{1}/_{3}m N C_{RMS}²where mN = mN₀ = M.

Ideal gas law for one-mole gas

PV = RT.

∴ PV =

^{1}/_{3}m N C_{RMS}²or, RT =

^{1}/_{3}M C_{RMS}²or, C

_{RMS}² = 3RT/M∴ C

_{RMS}= √3RT/MProblem

Calculate the root mean square speed of oxygen gas at 27⁰C.

Solution

RMS velocity of a gas molecule

C

_{RMS}² = (3RT/M).

M = 32 gm mol⁻¹.

T = 27° C = (273+27)K = 300 K.

C

_{RMS}² = (3 × 8.314 × 10⁷ × 300)/(32)

∴ C

_{RMS}= 48356 cm sec⁻¹

Problem

Calculate the RMS speed of ammonia at N.T.P.

Solution

At N.T.P,

V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹

P = 1 atm = 101325 Pa

M = 17 × 10⁻³ Kg mol⁻¹.

C

= (3 × 101325 × 22.4 × 10⁻³) /(17 × 10⁻³)

∴ C

V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹

P = 1 atm = 101325 Pa

M = 17 × 10⁻³ Kg mol⁻¹.

C

_{RMS}² = 3RT/M= (3 × 101325 × 22.4 × 10⁻³) /(17 × 10⁻³)

∴ C

_{RMS}= 632 m sec⁻¹#### Compare RMS velocity of hydrogen and oxygen

C

Hence at a given temperature,

(C

= 32/2 = 16

∴ RMS velocity of oxygen = 4 × RMS velocity of hydrogen.

_{RMS}² = 3RT/MHence at a given temperature,

(C

_{RMS}of hydrogen)²/(C_{RMS}of oxygen)² = M_{O₂}/M_{H₂}= 32/2 = 16

∴ RMS velocity of oxygen = 4 × RMS velocity of hydrogen.

#### The kinetic energy of the gas molecules

The average kinetic energy defined as,

Ä’ =

Kinetic gas equation

PV =

or, PV = ²/

or, PV = ²/

Ideal gas law for 1-mole gas

PV = RT and N = N₀.

∴ RT = ²/

or, Ä’ = ³/₂ (R/N₀) × T

∴ Ä’ = ³/₂ kT

k = R/N

= 1.38 × 10

Total kinetic energy for 1 mole of the gas molecules

E

= ³/₂ RT

Ä’ =

^{1}/₂ m C_{RMS}².Kinetic gas equation

PV =

^{1}/_{3}m N C_{RMS}²or, PV = ²/

_{3}N ×^{1}/_{2}(m C_{RMS}²)or, PV = ²/

_{3}N Ä’.Ideal gas law for 1-mole gas

PV = RT and N = N₀.

∴ RT = ²/

_{3}N₀ Ä’or, Ä’ = ³/₂ (R/N₀) × T

∴ Ä’ = ³/₂ kT

k = R/N

_{0}= Boltzmann constant of gas= 1.38 × 10

^{-23}J K⁻¹.Total kinetic energy for 1 mole of the gas molecules

E

_{Total}= N₀ × (Ä’)= ³/₂ RT

Average kinetic energy dependent on temperature only and average kinetic energy independent of the nature of the gas.

Problem

Calculate the kinetic energy of translation of 8.5 gm ammonia at 27⁰C.

Solution

Kinetic energy for 1 mole of the gas molecule,

E

= (3/2)(2 × 300 K)

= 900 cal mol⁻¹

8.5 gm ammonia = (8.5/17) mol

= 0.5 mol

The kinetic energy of 8.5 gm ammonia at 27⁰C

= (0.5 × 900) calories

= 450 calories

E

_{total}= (3/2)RT.= (3/2)(2 × 300 K)

= 900 cal mol⁻¹

8.5 gm ammonia = (8.5/17) mol

= 0.5 mol

The kinetic energy of 8.5 gm ammonia at 27⁰C

= (0.5 × 900) calories

= 450 calories