wu3c7 by Adel Khamis

8/14/2019 wu3c7 by Adel Khamis

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Unit three -77- Chapter Seven

The kinetic theory of gases

and some of its applications

Kinetic theory of gases:

1. A gas is composed of perfectly elastic spheres called ..

2. The intermolecular distances are that the volume of the gass

molecules is negligible compared with the volume occupied by the gas

itself.

3. The intermolecular forces are . due to ., so they are

negligible.

4. Gas molecules are in continuous random motion called . and

during its motion it is colliding and ..

5. The collisions between the molecules are considered as perfectly elastic

collisions, so the kinetic energy of the gas molecules . before

and after the collision, and the velocity before collision .. the

velocity after collision.

6. Gas in thermal equilibrium with .. of container.

The mole:

Definition:

..

..

One mole of any substance contain the same number of molecules, and that

number is known as .

Work Sheet 2008/2009


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Avogadros law:

Statement:

....

According to Avogadros number and Avogadros law, we can conclude that

one mole of any gas occupy the (same / different) volume.

The mole of any gas occupy . liters at S.P.T

Calculate the number of molecules, and the volume occupied by 0.1 kg of

oxygen [o16] given that NA = 6.023x1023

....

Density:

1. Definition:

2. Rule:

3. For certain amount of gas contain N molecules and the mass of each

molecule is m, the total mass of the gas equal:.

4. The density of the gas (in terms of mass of molecule) =

Example:

Given that the density of hydrogen gas is 0.09 kg/m3 find the mass of

hydrogen molecule, if 1020 molecules occupy 100 cm3.


.........

............

=


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..

[9x10-26kg]

Pressure:

1. Pressure (definition):

2. Pressure (rule):

3. Momentum equal (rule): .

4. Force (Definition):

5. Force (rule): .

Considering cubic vessel of lenght (l) contains certain number of molecules

of gas:

6. Linear Momentum of one molecule before impact = .

7. Linear Momentum of one molecule after impact = .

8. The change in linear momentum = .

9. The change in momentum delivered to the wall =

.

10. Force of one molecule = =

11. From the law of velocity, t = = .

12. From (10 & 11) Force = ...

13. For the total number of molecules (N), F = ..

14. From (2 & 13) P = .



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15. The area of the cube base = , and the volume of cube

.

16. From (14 & 15) P = .

17. Generally, in the three dimension: V2 = . + .. +

., and since Vx2 = Vy

2 = Vz2, therefore V2 = ..

18. Therefore: Vx2 = .

19. From (16 & 18) P =

20. Density = .

21. From (19 & 20) P =

22. Velocity =

Where: v is called ..

23. Impulse = = =

N.B.:

Number of collision per unit time can be calculated from the relation

.



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The Scientific Concept of Temperature:

The number of molecules in one mole of any gas is constant and called

.. (NA = 6.023x1023 molecules)

Macroscopic theory Microscopic theory

For any amount of gas

PVol = . PVol = .

For one mole of gas

N = .

n = .

PV = . PV = .. = ....

NAmv2 =

..............

..............2=mv

mv2 = .

(where K called ..)

Kinetic energy Temperature

mv2 = ..............

..............2=v

K.E = ..

..........

..........=v

K.E depends on Velocity depends on ..

and .

Where:

R is called (R = 8.31, and its

unit is )

K is called ............

........=K (K = 1.38 x 10-

23 J/K)



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N.B.:

The number of molecules in one mole of any gas is constant and called

. (NA = 6.023x10

23

molecules)

One mole of any gas occupies .. liter at SPT

It is (possible / impossible) for two gases to occupy the same volume at

the same time. (Give reason)

The total pressure of two gases at the same container =

.

Give reasons:

R.M.S.S does not affect by pressure.

There is no Hydrogen gas in the earths atmosphere.

Examples:

1. If the density of the hydrogen at normal temperature and pressure (NTP)

is 0.089 kg/m3, calculate the root mean square speed of its molecules.

.

.

.

.

[1847.87 m/s]



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2. Calculate the number of molecules per unit volume of oxygen at NTP

using the following data:

v = 4.62 x 102 m/s

m = 52.8 x 10-27 kg.

.

.

.

.

[2.7x1025]

3. Nitrogen gas in a container at NTP. Calculate the root mean square speed

of its molecules given that the mass of one Nitrogen mole is 0.028 Kg

and occupies 22.4 liters at NTP and that Avogadros number equals 6.02

x 1023 Molecules, Boltzmanns constant equals 1.38 x 10-23 J/K.

[493 m/s]


wu3c7 by Adel Khamis

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