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Daltons Law

by Dasa . 27 No 2021

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

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