Thermal Expansion of Gases

1. Relationship between volume and temperature at constant pressure (Charles's law)

Charles's law states that the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, as long as the pressure remains constant.

Formula:

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

Example 1:

A gas has a volume of 30 cm³ at 30°C. What will the volume be at 60°C?

Solution:

Convert temperatures to Kelvin:

$T_1 = 30 + 273 = 303$ K

$T_2 = 60 + 273 = 333$ K

$V_1 = 30$ cm³

$V_2 = \frac{V_1 \times T_2}{T_1} = \frac{30 \times 333}{303} = 32.97\text{ cm}^3$

2. Relationship between pressure and volume at constant temperature (Boyle's law)

Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to its volume, when temperature is constant.

Formula:

$P_1V_1 = P_2V_2$

Example 2:

A gas has a volume of 50 cm³ at 1.2 × 10⁵ Pa. What is the volume when the pressure increases to 2.0 × 10⁵ Pa?

Solution:

$V_2 = \frac{P_1 \times V_1}{P_2} = \frac{1.2 \times 10^5 \times 50}{2.0 \times 10^5} = 30\text{ cm}^3$

3. Relationship between pressure and temperature at constant volume (pressure law)

Pressure law states that the pressure of a fixed mass of gas is directly proportional to its absolute temperature, when volume is constant.

Formula:

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

Example 3:

The pressure of a gas is 1.0 × 10⁵ Pa at 0°C. What is the pressure at 150°C?

Solution:

$T_1 = 0 + 273 = 273$ K

$T_2 = 150 + 273 = 423$ K

$P_2 = \frac{P_1 \times T_2}{T_1} = \frac{1.0 \times 10^5 \times 423}{273} = 1.54 \times 10^5\text{ Pa}$

4. General gas law (combined gas law)

By combining the three laws, we get the general gas law:

Formula:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

This equation helps to solve problems where pressure, volume, and temperature all change.

1. Absolute zero: Lowest possible temperature, -273°C = 0 K
2. Kelvin is the SI unit of temperature.
3. 0°C = 273 K and 100°C = 373 K

Conversions:

1. °C to K: K = °C + 273
2. K to °C: °C = K - 273

Example 4:

Convert 65°C to Kelvin:

$T(K) = 65 + 273 = 338\text{ K}$

1. Standard temperature: 0°C or 273 K
2. Standard pressure: 1 atmosphere = 101300 Pa or 760 mmHg

1. Land and sea breezes: Caused by unequal heating and cooling of land and water.
2. Piston engines: Work due to expansion of hot gases.
3. Guns and bullets: Bullets are fired due to rapid gas expansion.