Gas Law – SS1 Chemistry Lesson Note (Second Term)

Gas laws describe how the volume, pressure and temperature of a fixed mass of gas are related. Boyle’s law links volume and pressure at constant temperature, and Charles’ law links volume and temperature at constant pressure. This note also covers the general and ideal gas equations and the laws of Dalton, Avogadro, Gay Lussac and Graham.

Behavioural objectives

By the end of the lesson, students should be able to:

• state Boyle’s law, Charles’ law, Dalton’s law of partial pressure, Avogadro’s law, Gay Lussac’s law of combining volumes and Graham’s law of diffusion;
• explain Boyle’s law and Charles’ law using the kinetic theory;
• use the general gas equation and the ideal gas equation to solve calculations;
• calculate partial pressures of gases in a mixture;
• use Gay Lussac’s law and Graham’s law to solve calculations.

What is Boyle’s law?

Boyle’s law states that the volume of a fixed mass of gas is inversely proportional to the pressure, provided the temperature remains constant.

Mathematical form of Boyle’s law

V ∝ 1/P
V = k/P
PV = k

Hence, P₁V₁ = P₂V₂

Boyle’s law can be represented graphically as shown below.

The graph shows that if the pressure is doubled, the volume is reduced to half its former value, and if the pressure is halved, the volume is doubled.

How does the kinetic theory explain Boyle’s law?

When the volume of a fixed mass of gas is decreased, the molecules of the gas collide with the walls of the container more frequently. This gives rise to an increase in pressure. However, if the molecules are farther apart, the number of collisions per unit time decreases, resulting in a decrease in pressure.

What is Charles’ law?

Charles’ law states that the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature on the Kelvin scale.

Mathematical form of Charles’ law

V ∝ T
V = kT
V/T = k

Hence, V₁/T₁ = V₂/T₂

The graphical representation of Charles’ law is shown below.

How does the kinetic theory explain Charles’ law?

When a given gas is heated at constant pressure, the molecules gain more kinetic energy and move faster. They collide with one another and with the walls of the container more frequently. To keep the number of collisions on the walls of the container the same (that is, to keep the pressure constant), the volume of the gas increases.

How do we solve calculations based on Boyle’s and Charles’ laws?

Example 1 (Boyle’s law)

200 cm³ of a gas has a pressure of 510 mmHg. What will be its volume if the pressure is increased to 780 mmHg, assuming there is no change in temperature?

Solution:

V₁ = 200 cm³, P₁ = 510 mmHg, P₂ = 780 mmHg, V₂ = ?

Using the expression for Boyle’s law:

P₁V₁ = P₂V₂

V₂ = P₁V₁/P₂ = (510 mmHg × 200 cm³)/780 mmHg = 130.769 = 131 cm³

Example 2 (Charles’ law)

A certain mass of a gas occupies 300 cm³ at 35°C. At what temperature will its volume be reduced by half, assuming its pressure remains constant?

Solution:

V₁ = 300 cm³, T₁ = 35°C = (35 + 273) K = 308 K, V₂ = V₁/2 = 300/2 = 150 cm³, T₂ = ?

Using the formula for Charles’ law:

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁ = (150 cm³ × 308 K)/300 cm³ = 154 K

What is the general gas equation?

Boyle’s and Charles’ laws are combined into a single expression known as the general gas equation, which can be expressed mathematically as:

P₁V₁/T₁ = P₂V₂/T₂

What is the ideal gas equation?

This equation states that for an ideal gas, PV/T is a constant. For one mole of gas, PV/T = R (R = molar gas constant), so:

PV = RT

For n moles of a gas, the equation becomes:

PV = nRT

Calculations

Example 1. What is the volume at s.t.p. of a fixed mass of gas that occupies 700 cm³ at 25°C and 0.84 × 10⁵ N m^-2 pressure?

Solution:

T₁ = 273 K, P₁ = 1.01 × 10⁵ N m^-2, T₂ = 25°C = (25 + 273) K = 298 K, P₂ = 0.84 × 10⁵ N m^-2, V₂ = 700 cm³, V₁ = ?

Using the general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

V₁ = P₂V₂T₁/(P₁T₂) = (0.84 × 10⁵ N m^-2 × 700 cm³ × 273 K)/(1.01 × 10⁵ N m^-2 × 298 K) = 533.337 = 533 cm³

Example 2. Calculate the number of moles present in a certain mass of gas occupying 6.5 dm³ at 3 atm and 15°C (R = 0.082 atm dm³ K^-1 mol^-1).

Solution:

V = 6.5 dm³, P = 3 atm, T = 15°C = (15 + 273) K = 288 K, n = ?

Using PV = nRT:

n = PV/RT = (3 atm × 6.5 dm³)/(0.082 atm dm³ K^-1 mol^-1 × 288 K) = 0.8257

Number of moles = 0.83 mole

What is Dalton’s law of partial pressure?

Dalton’s law states that in a mixture of gases which do not react chemically together, the total pressure exerted by the mixture is equal to the sum of the partial pressures of the individual gases that make up the mixture.

Mathematically, the law can be expressed as:

P_total = P_A + P_B + P_C + … + P_n

Here P_total is the total pressure of the mixture, and P_A, P_B and P_C are the partial pressures exerted separately by the individual gases A, B and C that make up the mixture.

The pressure each constituent gas exerts is called its partial pressure. The partial pressure of gas A is:

P_A = (number of moles of gas A / total number of moles of gas in the mixture) × P_total

That is, P_A = n_A/(n_A + n_B + n_C) × P_total

If the gas is collected over water, it is likely to be saturated with water vapour, and the total pressure becomes:

P_total = P_gas + P_{water vapour}
P_gas = P_total – P_{water vapour}

Calculation on Dalton’s law

A gaseous mixture containing 64 g of O₂ and 70 g of N₂ exerts a total pressure of 1.80 atm. What is the partial pressure exerted by oxygen in the mixture?

Solution:

Molar mass of O₂ = 16 × 2 = 32 g mol^-1
Molar mass of N₂ = 14 × 2 = 28 g mol^-1

Number of moles of O₂ = 64 g / 32 g mol^-1 = 2.0 mole
Number of moles of N₂ = 70 g / 28 g mol^-1 = 2.5 mole

Total number of moles of gases in the mixture = 2.0 + 2.5 = 4.5 mole

Partial pressure of O₂ = (2.0/4.5) × 1.80 atm = 0.80 atm

What is Avogadro’s law?

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. This means that 1 mole of any gas at s.t.p. has a volume of 22.4 dm³.

What is Gay Lussac’s law of combining volumes?

This law states that when gases react, they do so in volumes which are in simple ratios to one another and to the volumes of the products, if the products are gaseous, provided that the temperature and pressure remain constant.

Calculation on Gay Lussac’s law

Calculate the volume of oxygen required to burn 500 cm³ of methane completely.

Solution:

The equation for the reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

By Gay Lussac’s law, 1 volume of CH₄ requires 2 volumes of O₂ for complete combustion.

Therefore, 1 cm³ of CH₄ requires 2 cm³ of O₂, and 500 cm³ of CH₄ will require X cm³ of O₂.

X = (500 cm³ × 2 cm³)/1 cm³ = 1000 cm³

Thus, 1000 cm³ of O₂ is required.

What is Graham’s law of diffusion?

Graham’s law states that the rate of diffusion of a gas is inversely proportional to the square root of its density at constant temperature and pressure.

Mathematically,

R ∝ 1/√d
R = k/√d, where k is a constant

Comparing the rates of diffusion of two gases:

R₁/R₂ = √d₂/√d₁

In terms of relative molecular mass, M:

R ∝ 1/√M

For two gases,

R₁/R₂ = √M₂/√M₁

The rate of diffusion is the reciprocal of time, R = 1/t. That is,

R₁/R₂ = t₂/t₁

From the inverse relationship, we can deduce that the less dense a gas is, the higher its rate of diffusion, and the denser a gas is, the lower its rate of diffusion.

Calculation on Graham’s law

A given volume of SO₂ diffuses in 60 seconds. How long will it take the same volume of CH₄ to diffuse under the same conditions? (SO₂ = 64, CH₄ = 16)

Solution:

Using the expression:

t₂/t₁ = √M₂/√M₁

t₂ = (√M₂/√M₁) × t₁ = (√16/√64) × 60 seconds = 30 seconds

Evaluation

1. State Boyle’s law and Charles’ law.
2. Using the kinetic theory, explain why the volume of a gas increases when it is heated at constant pressure.
3. State Dalton’s law of partial pressure and write the expression for the partial pressure of gas A in a mixture.
4. What volume does 1 mole of any gas occupy at s.t.p.?
5. Write the relationship between the rates of diffusion of two gases and their relative molecular masses.

Assignment

1. A given mass of nitrogen occupies 0.12 dm³ at 60°C and 1.01 × 10⁵ N m^-2. Find its pressure at the same temperature if its volume is changed to 0.24 dm³.
2. 272 cm³ of CO₂ was collected over water at 15°C and 782 mmHg pressure. Calculate the volume of the dry gas at s.t.p. (saturated vapour pressure of water at 15°C is 12 mmHg).
3. State Graham’s law of diffusion.
4. Under the same conditions of temperature and pressure, hydrogen diffuses 8 times as fast as gas Y. Calculate the relative molecular mass of Y.
5. State the following rule and principle: (a) Hund’s rule of maximum multiplicity, (b) Aufbau principle.
6. Write the electronic configuration of (a) oxide ion, (b) aluminium ion, (c) potassium, (d) phosphorus.