# Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure (1807):

This law states that, “The total pressure developed by a mixture of gases is equal to the sum of the partial pressure developed by the individual gases”.

I. e. $P = p_1 + p_2 + p_3 + p_4 + \cdots$

Where P is the total pressure and $p_1, p_2, p_3, p_4 \cdots$ are the partial pressures of the different non-reacting gases. The partial pressure may be defined as the pressure developed non-reacting gases.

The partial pressure may be defined as the pressure developed by a gas when it is present alone in the same container at the same temperature.

I. e.

Conversion of factors and constants $\dfrac{F- 32}{9} = \dfrac{C}{5} \\ TK = (t^0C + 273) \\[3mm] 1 atm = 76 cm Hg = 760 mm Hg \\[3mm] = 760 torr \\[3mm] = 101324 Pa = 101325 Nm^{-2} \\[3mm] = 1.011325 bar = 14.7 Ib in^{-2} \\[3mm] 1J = 10^7 erg = 0.239 cal \\[3mm] 1 g mole = 22.4 L \text{of a gas at STP} \\[3mm] R = 0.0821 L atm K^{-1} mol^{-1} \\[3mm] =8.314 JK^{-1} mol^{-1} \\[3mm] 8.314 \times 10^7 erg K^{-1} mol^{-1} \\[3mm] = 1.987 = 2 cal K^{-1} mol^{-1}$ $\text{Partial pressure} = \dfrac{\text{Initial volume} \times \text{Initial pressure}}{\text{Total pressure}}$ $\% \text{of gas in mixture} = \dfrac{\text{Partial pressure}}{\text{total pressure}} \times 100$ $\text{in terms of number of moles},$ $P_A = \dfrac{\text{Moles of A}}{\text{Total moles}} \times P_{mix.}$ $P_{mix.} = \dfrac{(n_1 + n_2 + n_3 \cdots )}{V}RT$ $\left( \text{as} n = \dfrac{w}{m} \right)$ Where $n_1, n_2, n_3 \cdots$ are the number of the moles of different gases present in the mixture.

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