Gas laws
Dalton's law of partial pressure

Dalton's law of partial pressure states that the total pressure exerted by a mixture of gases which do not react chemically is equal to the sum of the individual partial pressures exerted by each gas in the mixture.

$$ P_T = P_1 + P_2 + P_3 + --- P_n $$ $$ \text{Where, }P_T = \text{total pressure} $$ $$ \text{The partial pressure of any gas is} $$ $$ P_1 = \text{mole fraction} × P_T $$ $$ \text{mole fraction}= \frac{n_1}{n_1 + n_2 + --- n_n} $$ $$ n = \text{number of moles of gases} $$ $$ \text{Partial pressure} = \frac{n_1}{n_T} × P_T $$ $$\text{where } n_T = n_1 + n_2 + n_3 --- + n_n $$

Note: When a gas is collected over water, it is likely to saturated with water vapour and the total pressure is given by

$$ P_{total} = P_{gas} + P_{\text{water vapour}} $$ $$ P_{gas} = P_{total} - P_{\text{water vapour}} $$

The vapor pressure of water is usually given as S.V.P of water at that temperature.

Calculations

Example 1: The volume of a sample of methane, collected over water at a temperature of 12°C and 700mmHg was 30cm³. Calculate the volume of the dry gas at S.T.P [Saturated vapour pressure of water at 12°C is 10mmHg] (WAEC)

Solution

$$ P_T = 700mmHg $$ $$ P_{\text{water vapour}} = 10mmHg $$ $$ P_1 = 700 - 10 = 690mmHg $$ $$ T_1 = 12 + 273 = 285K $$ $$ V_1 = 30cm³ $$ $$ P_2 = 760mmHg $$ $$ V_2 = ? $$ $$ T_2 = 273K $$ $$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$ $$ V_2 = \frac{P_1V_1T_2}{P_2T_1} $$ $$V_2 = \frac{690 × 30 × 273}{760 × 285} $$ $$ V_2 = 26.09cm³ $$

Example 2 : A certain mass of gas collected over water at 16°C and 745mmHg pressure has a volume of 40cm³. Calculate the volume when it is dry at s.t.p. [S.V.P of water at 16°C = 13.5mmHg] (WAEC)

Solution

$$ P_T = 745mmHg $$ $$ P_{\text{water vapour}} = 13.5mmHg $$ $$ P_1 = 745 - 13.5 = 731.5mmHg $$ $$ V_1 = 40cm³ $$ $$ T_1 = 16°C + 273 = 289K $$ $$ P_2 = 760mmHg $$ $$ T_2 = 273K $$ $$ V_2 = ? $$ $$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$ $$ V_2 = \frac{P_1V_1T_2}{P_2T_1} $$ $$ V_2 = \frac{731.5 × 40 × 273}{760 × 289} $$ $$ V_2 = 36.37cm³ $$

Example 3: An enclosed vessel 5.6g of oxygen and 13g of chlorine at STP. What will be the partial pressure of Chlorine, if the temperature is raised to 200°C. (O = 16, Cl = 35.5)

Solution

$$ \text{Mass of oxygen} = 5.6g $$ $$ \text{Mass of Chlorine} = 13g $$

Since Volume is constant, we use pressure law.

$$ \text{At STP} $$ $$ P_1 = 760mmHg $$ $$ T_1 = 273K $$ $$ P_2 = ? $$ $$ T_2 = 200°C + 273 = 473K $$ $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ $$ {P_2} = \frac{P_1T_2}{T_1} $$ $$ P_2 = \frac{760 × 473}{273} $$ $$ P_2 = 1316.78mmHg $$

The partial pressure of Chlorine is calculated by:

$$ P_{Cl_{2}} = \frac{n_{cl_2}}{n_{total}} × P_T $$ $$ \text{n}Cl_2 = \frac{mass}{\text{molar mass}} $$ $$ n_{Cl_2} = \frac{13}{35.5 × 2} $$ $$ = \frac{13}{71} $$ $$ n_{Cl_2} = 0.183mole $$

The number of moles of oxygen is given by:

$$ n_{O_2} = \frac{mass}{\text{molar mass}} $$ $$ n_{O_2} = \frac{5.6}{16×2} $$ $$ n_{O_2} = 0.175 mole $$

The total number of moles is the sum of that of oxygen and chlorine

$$ n_T = n_{Cl_2} + n_{O_2} $$ $$ n_T = 0.183 + 0.175 $$ $$ n_T = 0.358 $$ $$ P_{Cl_2} = \frac{n_{Cl_2}}{n_T} × P_T $$ $$ P_{Cl_2} = \frac{0.183}{0.358} × 1316.78 $$ $$ P_{Cl_2} = 673.10mmHg $$

Gas Equation Calculator
















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

  2. For 2: \( T_2 = \frac{P_2V_2T_1}{P_1V_1} \)
  3. K = °C + 273
  4. \(P_{gas} = \frac{n_{gas}}{n_{\text{total gases}}} × P_T \)
This calculator is programmed with the general gas equation. Use e for exponential values. Calculate for P1 before inputting

Summary