Kinetic Theory of Gases

1. Gases contain a large number of molecules which are in continuous random motion.
2. Molecules of gases are far apart such that the force of attraction between them is negligible. The kinetic energy of the molecules is directly proportional to the absolute temperature, i.e., an increase in temperature causes an increase in kinetic energy.
3. The volume of individual gas molecules is negligible compared to the volume of the container.
4. The pressure of a gas inside a container is due to collisions between the gas molecules and the walls of the container.
5. Collisions of gas molecules are perfectly elastic, meaning kinetic energy is conserved (kinetic energy before collision equals kinetic energy after collision).

Ideal gases

Ideal gases obey all assumptions of the kinetic theory of gases and follow the ideal gas equation:

$PV = nRT$

• P: Pressure of the gas
• V: Volume of the gas
• n: Number of moles of gas
• R: Universal gas constant
• T: Temperature

Real gases

Real gases do not obey all assumptions of the kinetic theory. For example:

• The volume of individual gas molecules cannot always be considered negligible.
• Forces of attraction between gas molecules may not be negligible.

Charles's law

States that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

Boyle's law

States that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

Graham's law of diffusion

States that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its density.

$\text{Rate} \propto \frac{1}{\sqrt{\rho}}$

Dalton's law of partial pressure

States that the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases:

$P_T = P_A + P_B$

The ideal gas equation is:

$PV = nRT$

For one mole of gas:

$PV = RT$

The universal gas constant $R$ has values:

• $R = 0.0821$ L·atm·mol$^{-1}$·K$^{-1}$
• $R = 8.314$ J·mol$^{-1}$·K$^{-1}$

At high pressure and low temperature, real gases deviate from ideal behaviour due to:

• Intermolecular forces
• Finite molecular volume

Van der Waals equation accounts for these deviations:

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

• a: Attraction between molecules
• b: Volume correction factor

1. Determination of molar mass of gases
2. Calculation of gas densities
3. Prediction of gas behavior under different conditions

Example 1

Question: A gas occupies a volume of 3.25 dm³ at a pressure of 1.01 × 10⁵ Nm⁻² and a temperature of 298 K. If the volume is compressed to 1.88 dm³, calculate the final pressure, assuming the temperature remains constant.

Solution:

Given:

• Initial volume, $V_1 = 3.25$ dm³
• Final volume, $V_2 = 1.88$ dm³
• Initial pressure, $P_1 = 1.01 \times 10^5$ Nm⁻²

Using Boyle's Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$(1.01 \times 10^5) \times 3.25 = P_2 \times 1.88$

Calculate:

$P_2 = \frac{1.01 \times 10^5 \times 3.25}{1.88} = 1.74 \times 10^5 \text{ Nm}^{-2}$

Final Answer: $P_2 = 1.74 \times 10^5$ Nm⁻²

Example 2

Question: Calculate the volume occupied by 13.7 g of chlorine gas at 45°C and a pressure of 745 mmHg. Use $R = 0.0821$ L·atm·mol$^{-1}$·K$^{-1}$.

Solution:

Given:

• Mass, $m = 13.7$ g
• Molar mass of $\text{Cl}_2$, $M = 71$ g·mol$^{-1}$
• Temperature, $T = 45 + 273 = 318$ K
• Pressure, $P = 745 / 760 = 0.980$ atm

From the ideal gas equation:

$PV = nRT$

$V = \frac{nRT}{P}$

Calculate the number of moles:

$n = \frac{m}{M} = \frac{13.7}{71} = 0.193 \text{ mol}$

Substitute into the equation:

$V = \frac{0.193 \times 0.0821 \times 318}{0.980} = 5.14 \text{ L}$

Final Answer: $V = 5.14$ L

Example 3

Question: What is the temperature of oxygen gas if its density is $1.00$ g·dm$^{-3}$, pressure is 745 mmHg, and $R = 0.0821$ L·atm·mol$^{-1}$·K$^{-1}$?

Solution:

Given:

• Density, $\rho = 1.00$ g·dm$^{-3}$
• Pressure, $P = 745 / 760 = 0.980$ atm
• Molar mass of oxygen, $M = 32$ g·mol$^{-1}$

From the ideal gas equation:

$PV = nRT$

Rewriting in terms of density:

$T = \frac{PM}{\rho R}$

Substitute the known values:

$T = \frac{0.980 \times 32}{1.00 \times 0.0821} = 382.5 \text{ K}$

Final Answer: $T = 382.5$ K

Example 4

Question: A gas takes 54.4 s to effuse through a porous plug. Under the same conditions, oxygen gas takes 36.5 s. Calculate the molar mass of the gas.

Solution:

Using Graham's Law:

$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$

Where $r_1$ and $r_2$ are rates of effusion, inversely proportional to time:

$\frac{t_2}{t_1} = \sqrt{\frac{M}{32}}$

Substitute the values:

$\frac{54.4}{36.5} = \sqrt{\frac{M}{32}}$

Square both sides:

$\left(\frac{54.4}{36.5}\right)^2 = \frac{M}{32}$

$M = 32 \times \left(\frac{54.4}{36.5}\right)^2 = 71.0 \text{ g·mol}^{-1}$

Final Answer: $M = 71.0$ g·mol$^{-1}$

Example 5

Question: A sample of hydrogen gas has a volume of 5.00 L at a pressure of 1.00 atm and a temperature of 300 K. Calculate the volume when the pressure is increased to 2.00 atm, keeping the temperature constant.

Solution:

Given:

• Initial volume, $V_1 = 5.00$ L
• Initial pressure, $P_1 = 1.00$ atm
• Final pressure, $P_2 = 2.00$ atm
• Temperature remains constant.

Using Boyle's Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$(1.00 \times 5.00) = (2.00) \times V_2$

Solving for $V_2$:

$V_2 = \frac{1.00 \times 5.00}{2.00} = 2.50 \text{ L}$

Final Answer: $V_2 = 2.50$ L

Example 6

Question: A gas sample has a volume of 4.50 L at 2.00 atm pressure and a temperature of 300 K. What will be the volume of the gas at 1.00 atm and 350 K?

Solution:

Given:

• Initial volume, $V_1 = 4.50$ L
• Initial pressure, $P_1 = 2.00$ atm
• Initial temperature, $T_1 = 300$ K
• Final pressure, $P_2 = 1.00$ atm
• Final temperature, $T_2 = 350$ K

Using the Combined Gas Law:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Substitute the known values:

$\frac{2.00 \times 4.50}{300} = \frac{1.00 \times V_2}{350}$

Solving for $V_2$:

$V_2 = \frac{2.00 \times 4.50 \times 350}{300 \times 1.00} = 7.00 \text{ L}$

Final Answer: $V_2 = 7.00$ L

Example 7

Question: A gas occupies a volume of 10.0 L at a temperature of 273 K and a pressure of 1.00 atm. What will be the pressure when the gas is heated to 313 K, assuming the volume remains constant?

Solution:

Given:

• Initial volume, $V_1 = 10.0$ L
• Initial pressure, $P_1 = 1.00$ atm
• Initial temperature, $T_1 = 273$ K
• Final temperature, $T_2 = 313$ K
• Volume remains constant.

Using Gay-Lussac's Law:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Substitute the known values:

$\frac{1.00}{273} = \frac{P_2}{313}$

Solving for $P_2$:

$P_2 = \frac{1.00 \times 313}{273} = 1.15 \text{ atm}$

Final Answer: $P_2 = 1.15$ atm

Example 8

Question: A gas has a density of 1.10 g/L at 25°C and 1.00 atm pressure. Calculate the molar mass of the gas.

Solution:

Given:

• Density, $\rho = 1.10$ g/L
• Pressure, $P = 1.00$ atm
• Temperature, $T = 25°C = 298$ K
• $R = 0.0821$ L·atm·mol$^{-1}$·K$^{-1}$

Using the ideal gas law rearranged for molar mass:

$PV = nRT, \quad n = \frac{PV}{RT}$

The density is defined as:

$\rho = \frac{m}{V} = \frac{nM}{V}$

Rearranging for molar mass ($M$):

$M = \frac{\rho RT}{P}$

Substitute the known values:

$M = \frac{1.10 \times 0.0821 \times 298}{1.00} = 27.0 \text{ g/mol}$

Final Answer: $M = 27.0$ g/mol

Example 9

Question: A gas has a volume of 10.0 L at 1.00 atm and 273 K. What is the number of moles of gas present?

Solution:

Given:

• Volume, $V = 10.0$ L
• Pressure, $P = 1.00$ atm
• Temperature, $T = 273$ K
• $R = 0.0821$ L·atm·mol$^{-1}$·K$^{-1}$

Using the ideal gas equation:

$PV = nRT$

Solving for $n$:

$n = \frac{PV}{RT}$

Substitute the known values:

$n = \frac{1.00 \times 10.0}{0.0821 \times 273} = 0.446 \text{ mol}$

Final Answer: $n = 0.446$ mol

Example 10

Question: A balloon is filled with helium gas at 1.00 atm pressure and a volume of 20.0 L. If the balloon is compressed to 10.0 L, what is the new pressure of the gas?

Solution:

Given:

• Initial volume, $V_1 = 20.0$ L
• Final volume, $V_2 = 10.0$ L
• Initial pressure, $P_1 = 1.00$ atm

Using Boyle's Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$(1.00 \times 20.0) = P_2 \times 10.0$

Solving for $P_2$:

$P_2 = \frac{1.00 \times 20.0}{10.0} = 2.00 \text{ atm}$

Final Answer: $P_2 = 2.00$ atm