CBSE Class 10 Physics The Gas Laws Notes Set 17

Expansion of Solids, Liquids and Gases

Solids can expand in length (linear), in area (superficial) and in volume (cubical). Liquids and gases have only cubical expansion. The expansion of solids and liquids hardly depends on pressure because they are almost incompressible, but the volume of a gas depends on both temperature and pressure. Expansion is least in solids and greatest in gases. To study one factor, the other is kept constant, which gives Boyle's law and Charles' law.

Boyle's Law

At constant temperature, the volume of a given mass of gas is inversely proportional to its pressure, so PV = constant, or P₁V₁ = P₂V₂.

Example. A gas of 50 cm³ at 76 cm of mercury is expanded to 100 cm³ at constant temperature: P = 76 × 50 / 100 = 38 cm of mercury. For a closed tube with a mercury thread, the pressure of the trapped air is the atmospheric pressure plus the thread length when the open end is up, the atmospheric pressure alone when the tube is horizontal, and the atmospheric pressure minus the thread length when the open end is down.

Absolute Temperature

The Kelvin (absolute) scale has its zero at -273 °C, so T = t + 273. Zero kelvin is a theoretical limit never actually reached.

Charles' Law

At constant pressure, all gases expand by 1/273 of their volume at 0 °C for each °C rise in temperature. Thus V_t = V₀(1 + αt) with α = 1/273 per °C (the volume coefficient). Dividing two such equations gives V₁/T₁ = V₂/T₂: the volume is proportional to the absolute temperature.

At constant volume, pressure rises by 1/273 of its value at 0 °C for each °C, so P₁/T₁ = P₂/T₂. Examples: 600 cm³ at 27 °C becomes 600 × 400 / 300 = 800 cm³ at 127 °C; 75 cm of mercury at 27 °C becomes 100 cm of mercury at 127 °C.

To find the coefficient from two readings, write V₁ = V₀(1 + αt₁) and V₂ = V₀(1 + αt₂) and divide. For 500 cm³ at 27 °C and 600 cm³ at 87 °C: 5(1 + 87α) = 6(1 + 27α), so α = 1/273.

General Gas Equation and Universal Gas Constant

Combining both laws: PV/T = constant, so P₁V₁/T₁ = P₂V₂/T₂. Example: 200 cm³ at 75 cm and 27 °C becomes 200 × 75 × 500 / (300 × 100) = 250 cm³ at 100 cm and 227 °C.

One mole of any gas occupies 22.4 litres at S.T.P. (0 °C, 76 cm of mercury = 1.013 × 10⁵ Pa). Hence PV/T = (1.013 × 10⁵ × 22.4 × 10^-3)/273 = 8.31 J K^-1 mol^-1 = R, the universal gas constant. So PV = RT for one mole and PV = nRT for n moles.

Atmospheric pressure in SI units: P = hdg = 0.76 × 13.6 × 10³ × 9.8 = 1.013 × 10⁵ Pa, which is about 1.013 bar (1 bar = 10⁵ Pa).