Gas laws
Pressure law

Pressure law states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant. This means that as pressure increases, temperature increases as well and vice versa.

$$ P \propto T$$ $$ P = kT $$ $$ k = \frac{P}{T} $$ $$ \text{The mathematical expression is}; $$ $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ $$\text{Note: Always covert to kelvin scale} $$ $$ K = (°C + 273) $$
General gas equation

The general gas law states that the volume of a fixed mass of gas is directly proportional to its temperature and inversely proportional to its pressure. It is a combination of Boyle's law and Charles law.

$$ \frac{PV}{T} = K $$ $$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$

Solution

Example 1: At S.T.P. a certain mass of gas occupies a volume of 1200cm³. Find the temperature at which the gas occupies 820cm³ and has a pressure of 515mmHg.

Solution

Since the question said at Standard temperature and pressure, hence the initial pressure and temperature are both standard values

$$ P_1 = 760mmHg $$ $$ T_1 = 273k $$ $$ V_1 = 1200cm³ $$ $$ P_2 = 515mmHg $$ $$ T_2 = ? $$ $$ V_2 = 820cm³ $$ $$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$ $$ T_2 = \frac{P_2V_2T_1}{P_1V_1} $$ $$ T_2 = \frac{515 × 820 × 273}{760 × 1200} $$ $$ T_2 = 126.4k $$

Example 2: A given mass of gas occupies 850cm³ at 320K and 0.92 × 10⁵ N/m² pressure. Calculate the volume of the gas at S.T.P.

Solution

$$ P_1 = 0.92 × 10^5 N/m² $$ $$ V_1 = 850cm³ $$ $$ T_1 = 320K $$ $$ V_2 = ? $$ $$ T_2 = 273K $$ $$ P_2 = 1.01325 × 10^5 N/m²$$ $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$ $$ V_2 = \frac{P_1V_1T_2}{P_2T_1} $$ $$ V_2 = \frac{0.92 × 10^5 × 850 × 273}{1.01325 × 10 ^5 × 320} $$ $$ V_2 = 658.42 cm³ $$

Gas Equation Calculator









Formulas:
  1. For 1: \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)

  2. For 2: \( T_2 = \frac{P_2V_2T_1}{P_1V_1} \)
This calculator is programmed with the general gas equation. Use e for exponential values

Summary