Kinetic Theory Of Gases Assumptions

Kinetic Theory Of Gases Assumptions

Edited By Vishal kumar | Updated on Sep 26, 2024 11:04 AM IST

The Kinetic Theory of Gases provides a microscopic understanding of how gases behave, focusing on the motion of individual gas molecules. It rests on several assumptions that simplify the complex nature of gases, making it easier to predict their properties like pressure, temperature, and volume. The theory assumes that gas molecules are in constant, random motion, have negligible volume compared to the container, and experience perfectly elastic collisions. These molecules do not exert attractive or repulsive forces on one another, except during collisions.

In real life, the assumptions of kinetic theory explain common experiences. For example, when you inflate a balloon, the gas molecules inside are constantly moving and colliding with the balloon’s walls, creating pressure that keeps it inflated. Similarly, when you heat a pot of water, the temperature rises and increases the speed of gas molecules, which explains why heated air rises. These assumptions allow us to understand and predict the behaviour of gases in everyday situations like weather patterns, vehicle engines, and even respiration.

Assumption of Ideal Gases

The assumptions of ideal gases are foundational to the kinetic theory of gases and simplify the behaviour of gases to make calculations easier. These assumptions describe an "ideal" gas that perfectly follows the gas laws.

What is Ideal gas?

It is a hypothetical gas (which is not real gas), whose molecules occupy negligible space and have no interactions (Force of interaction is much less), and which consequently obeys the gas laws exactly.

So, the ideal gas does not exist in real, but for study, we take some assumptions to make the gas ideal and we can apply some laws which are only valid for ideal gases. These assumptions are

  1. The size of the molecules is negligible in comparison to intermolecular distance (10-9m)
  2. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses (It means that when they collide with each other, then there is no loss of energy during collision).
  3. The molecules of a given gas are all identical but these molecules are different than those of another gas.
  4. The volume of molecules is negligible in comparison to the volume of gas.
  5. Molecules of a gas move randomly in all possible directions with all possible velocities.
  6. The speed of gas molecules varies from zero and infinity.
  7. The gas molecules keep on colliding among themselves as well as with the walls of the containing vessel. These collisions are perfectly elastic (no loss of energy).
  8. The time spent in a collision between two molecules is negligible in comparison to time between two successive collisions (i.e., time required to travel means free path).
  9. The number of collisions per unit volume in a gas remains constant.
  10. No attractive or repulsive force acts between gas molecules.
  11. Gravitational attraction among the molecules is negligible due to extremely small masses and the very high speed of molecules.
  12. Molecules constantly collide with the walls of the container due to which their momentum changes. The change in momentum is transferred to the walls of the container and due to this Pressure is exerted by gas molecules on the walls of the container.
  13. The density of gas does not change at any point in the container.

Recommended Topic Video

Solved Examples Based on the Assumption of Ideal Gases

Example 1: Which is true for an ideal gas?

(1) Molecule of gas is identical spherical rigid and perfectly elastic point mass.

(2) There is always some attractive and repulsive force acting between gas molecules.

(3) The density of a gas is constant at all points of the container molecules

1) only 1

2) 1 and 3

3) 2 and 3

4) only 3

Solution:

True: For an ideal gas, the molecules are considered to be identical, spherical, rigid, and perfectly elastic point masses. This is a fundamental assumption of the ideal gas model.

False: An ideal gas assumes no intermolecular forces (no attractive or repulsive forces) between gas molecules. This is key to the ideal gas behaviour.

False: The density of an ideal gas can vary within the container depending on temperature and pressure. It is not necessarily constant at all points in the container.

Hence, the answer is the option (1)

Example 2: For an ideal gas which statement is not true?

1) It obeys Boyle's Law

2) It follows PV = RT

3) Molecules of gas are identical spherical rigid and perfectly elastic point masses.

4) It follows Vander Waal's equation.

Solution:

Rest all the statements are correct and till now you have learned all those statements. But ideal gas does not obey the Van der Waals equation. Let us understand this in brief.

Van der Waals equation:

$ \left(P+\frac{a}{V^2}\right)(V-b)=R T$

To account for the volume that a real gas molecule takes up, the van der Waals equation replaces V in the ideal gas law with (V-b), where v is the molar volume of the gas and b is the volume that is occupied by one mole of the molecules.

Here a and b depend on the gas and unlike an ideal gas, the real gas molecules do interact among themselves (attraction or repulsion)

\therefore It does not follow Vander Waal's equation.

Hence, the answer is the option (4).

Example 3: According to the kinetic theory of gases,
A. The motion of the gas molecules freezes at $0^{\circ} \mathrm{C}$.
B. The mean free path of gas molecules decreases if the density of molecules is increased.
C. The mean free path of gas molecules increases if the temperature is increased keeping pressure constant.

D. Average kinetic energy per molecule per degree of freedom is $\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$ (for monoatomic gases).
Choose the most appropriate answer from the options given below:

1) $A$ and $C$ only
2) $B$ and C only
3) A and B only
4) C and D only

Solution:

Mean free path, $\lambda=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{d}^2 \rho}$

The mean free path is directly proportional to temperature $(\mathrm{T})$ and inversely proportional to the density $(\rho)$

Hence, the answer is the option (2).

Example 4:

There are two identical chambers, completely thermally insulated from the surroundings. Both chambers have a partition wall dividing the chambers into two compartments. Compartment 1 is filled with an ideal gas and compartment 3 is filled with a real gas. Compartments 2 and 4 are vacuum. A small hole (orifice) is made in the partition walls and the gases are allowed to expand in a vacuum.

Statement 1: No change in the temperature of the gas takes place when the ideal gas expands in a vacuum. However, the temperature of real gas goes down (cooling) when it expands in a vacuum.

Statement 2: The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential.

1) Statement 1 is false and statement 2 is true

2) Statement 1 and statement 2 both are true statement 2 is the correct explation of statement 1.

3) Statement 1 is true and statement 2 is false.

4) Statement 1 and statement 2 both are true. But statement 2 is not the correct explanation of statement 1.

Solution:

In ideal gases, the molecules are considered point particles and for point particles, there is no internal excitation, no vibration and no rotation. For an ideal gas, the internal energy can only be translational kinetic energy, and for real gas, both kinetic and potential energy.

Example 5: For an ideal gas relation between pressure and volume is $P=K V$,, the coefficient of pressure expansion is

1) T

2) 1/T

3) 2T

4) 1/2T

Solution:

Co-efficient of Pressure Expansion
$
\beta=\frac{\Delta P}{P_0} \frac{1}{\Delta \theta}
$

wherein

$
\begin{aligned}
& P^{\prime}=P(1+\beta \Delta \theta) \\
& P^{\prime}=\text { Final pressure } \\
& P=K V \\
& P V=n R T \Rightarrow V=\frac{n R T}{P} \\
& \text { or } P=K \cdot \frac{n R T}{P} \Rightarrow P^2=(K n R) T
\end{aligned}
$

on differentiating we get

$
2 P \cdot \frac{d P}{d T}=k n R
$
Coefficient of pressure expansion $=\frac{1}{P} \frac{d P}{d T}=\frac{1}{P} \frac{K n R}{2 P}=\frac{K n R}{2(k n R) T}=\frac{1}{2 T}$

Hence, the answer is the option (4).

Summary

The Kinetic Theory of Gases explains gas behaviour through assumptions like constant molecular motion, negligible volume, and no intermolecular forces, forming the basis for the ideal gas model. An ideal gas is hypothetical, following laws like Boyle's law and PV = RT, with assumptions such as perfectly elastic collisions and no intermolecular attraction. However, real gases deviate under specific conditions, requiring corrections like the Van der Waals equation to account for molecular interactions and volume.

Specific Heat Of A Gas

25 Sep'24 03:34 PM

Mean free path

10 Sep'24 08:37 PM

Degree of freedom

10 Sep'24 08:32 PM

Mayer's Formula

20 Aug'24 02:20 PM

Kinetic Energy Of Ideal Gas

16 Aug'24 03:57 AM

Articles

Get answers from students and experts
Back to top