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

**of gases based upon the certain postulates which are supposed to be applicable to an Ideal gas.**

*kinetic theory**Assumptions of the kinetic theory of gases*

- 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.
- The molecules are moving in all directions with the verity of speeds. Some are very fast while others are slow.
- 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
energy or momentum of the molecules by this collision.*kinetic* - The 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, wall collision becomes more frequent and pressure is increased.
- Through the molecular velocity is constantly changing due to the intermolecular collision, the average
energy(Ñ”) 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.*kinetic*

*Root mean square velocity*

- Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds.

C

_{RMS}² = (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 C

_{RMS}at temperature T and pressure P.

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 C

_{x}, C

_{y, }and C

_{z}.

C₁² = C

_{x}² + C_{y}² + C_{z}²- 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

= 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

= 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 C

where C

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

= 2 mN {(N₁C₁² + N₂C₂² + N₃C₃² ..)/N}

= 2 mN C

_{RMS}²where C

_{RMS}² = 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 C

or, P × l³ =

∴ PV =

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

The other form of the equation is,

P =

∴ P =

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

_{RMS}²/lor, P × l³ =

^{1}/_{3}m N C_{RMS}²∴ PV =

^{1}/_{3}m N C_{RMS}²Here, l³ = volume of the cube contains gas molecules.

The other form of the equation is,

P =

^{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 also valid for any shape of the gas container.

*Root mean square velocity*

Let us apply the

In that case mN = mN₀ = M

PV = RT.

Hence from the

or, RT =

or, C

∴ C

**equation for a 1-mole ideal gas.***kinetic*In that case mN = mN₀ = M

**,***Ideal gas equation*PV = RT.

Hence from the

**gas equation, PV =***kinetic*^{1}/_{3}m N C_{RMS}²or, RT =

^{1}/_{3}m N C_{RMS}²or, C

_{RMS}² = 3RT/M∴ C

_{RMS}= √3RT/M- Thus root means square velocity depends on the molar mass(M) and temperature(T) of the gas.

*Average kinetic energy*

The average

Ä’ =

Again from the

PV =

or, PV = ²/

or, PV = ²/

For 1 mole ideal gas,PV = RT and N = N₀

Thus RT = ²/

or, Ä’ = ³/₂R/N₀T

∴ Ä’ = ³/₂ kT

Where k = R/N

Its value is 1.38 × 10

The total

E

**energy(Ä’) is defined as,***kinetic*Ä’ =

^{1}/₂ m C_{RMS}²Again from the

**gas equation,***kinetic*PV =

^{1}/_{3}m N C_{RMS}²or, PV = ²/

_{3}N ×^{1}/_{3}m C_{RMS}²or, PV = ²/

_{3}N Ä’For 1 mole ideal gas,PV = RT and N = N₀

Thus RT = ²/

_{3}N₀ Ä’or, Ä’ = ³/₂R/N₀T

∴ Ä’ = ³/₂ kT

Where k = R/N

_{0}and is known as the Boltzmann constant.Its value is 1.38 × 10

^{-23}JK⁻¹The total

**energy for 1 mole of the gas is,***kinetic*E

_{Total}= N₀ (Ä’) = ³/₂RT- Thus average

**energy is dependent on T only and it is not dependent on the nature of the gas.**

*kinetic**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

PV = 1/3 mN C

or, P = 1/3 × (m N/V) × C

**gas equation,***kinetic*PV = 1/3 mN C

_{RMS}²or, P = 1/3 × (m N/V) × C

_{RMS}²Putting the value we have,

P = (1/3)(10⁻²⁵ Kg×10²³/10⁻³ m³) × (10³ m sec⁻¹)²

∴ P = 0.333 × 10⁷ atm

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,

C

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

Thus,C

∴ C

C

_{RMS}² = (3RT/M)Here, M = 32 gm mol⁻¹, and T = 270 C = (273+27)K = 300 K.

Thus,C

_{RMS}² = (3 × 8.314 × 10^{7}erg mol⁻¹ K⁻¹ × 300 K)/(32 gm mol⁻¹)∴ C

_{RMS}= 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, C

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

∴ C

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

∴ C

_{RMS}= 632 m sec⁻¹

*Problem*- How the root means square velocity for oxygen compares with that of the hydrogen?

*Answer*We know that, C

Hence at a given temperature,

(C

= 32/2 = 16

That is C

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

(C

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

That is C

_{RMS}of O₂ = 4 × C_{RMS}of H₂.

*Problem*- Calculate the

**energy of translation of 8.5 gm NH₃ at 27⁰ C.**

*kinetic*

*Solution*We know that total

E

= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K)

= 900 cal mol⁻¹

Again 8.5 gm NH

Thus the

= 450 cal

**energy for 1 mole of the gas is,***kinetic*E

_{Total}= (3/2)RT= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K)

= 900 cal mol⁻¹

Again 8.5 gm NH

_{3}= (8.5/17) mol = 0.5 molThus the

**energy of 8.5 gm NH***kinetic*_{3}at 27⁰ C is, (0.5 × 900) cal= 450 cal

*Problem*- Calculate the RMS velocity of oxygen molecules having a

**energy of 2 K.cal mol⁻¹. At what temperature the molecules have this value of KE?**

*kinetic*

*Solution*T = 673.9K or 400.9⁰ C