Daltons Law

When we place two gases in a container, the total pressure exerted by the mixture is

= Sum of partial pressure of each gas

When we place 1 mol of Oxygen & 1 mol of Hydrogen in a container

$P_{total} = P_{O_{2}} + P_{H_{2}}$

The ideal gas law

$P * V = n * R * T$

$P = \dfrac{n * R * T} {V}$

Where, P - Pressure exerted by the gas in $inHg$ n - Amount of gas in $lb-mol$ R - Universal gas constant = 21.85 $\dfrac{inHg*ft^{3}}{lb-mol* ^oR}$ T - Absolute temperature of the gas in $^oR$ V - Volume of the gas in $ft^3$

Example:

At 68° F, consider placing 1 lb of Hydrogen gas ($H_{2}$) and 1 lb of Oxygen gas ($O_{2}$) in a container of 15 $ft^3$.

Click here to get atomic mass in g/mol, for various elements.

H = 1.008 gram/mol H2 = 2.016 gram/mol = 2.016 ÷ 454 lb/mol = 0.00444 lb/mol

O = 15.999 gram/mol O2 = 31.998 gram/mol = 31.998 ÷ 454 lb/mol = 0.07048 lb/mol

Then,

1 lb of $H_{2}$ = 1 ÷ 0.00444 mol = 225.225 mol = 225.225 ÷ 454 lb-mol = 0.496 lb-mol

1 lb of $O_{2}$ = 1 ÷ 0.07048 mol = 14.188 mol = 14.188 ÷ 454 lb-mol = 0.03125 lb-mol

T = 68° F = 68 + 459.67 ° R = 527.67° R

Partial Pressure exerted by H2

$P_{H_{2}} = \dfrac{0.496 * 21.85 * 527.67} {15}$ $= 381.25\,mmHg$

Partial Pressure exerted by O2

$P_{O_{2}} = \dfrac{0.03125 * 21.85 * 527.67} {15}$ $= 24.02\,mmHg$

Total Pressure exerted

= 381.25 + 24.02 = 405.27 mmHg