A student has a 2.66 L bottle that contains a mixture of O2 , N2 , and CO2 with a total pressure of 4.50 bar at 298 K . She knows that the mixture contains 0.297 mol N2 and that…

Question

asked Feb 8, 2021 170k views

1 Answer

Lester Buck · Answer 1 · 2021-02-13T11:19:31+0000

Answer:

The partial pressure of
O_(2) = 1.455 bar.

Step-by-step explanation:

Given:

Volume
V = 2.66 L

Total pressure
$P_{} = 4.50$ bar

Temperature of system
T = 298 K

Moles of nitrogen
n = 0.297 mole

Partial pressure
$P_{co_(2) } = 0.269$ bar

From ideal gas equation,

PV = nRT

Where
R = 8.314 * 10^(-2) (L .bar)/(K.mol) = gas constant

First finding partial pressure of nitrogen

$P_{N_(2) } = (nRT)/(V)$

$P_{N_(2) } = (0.297* 8.314 * 10^(-2) * 298)/(2.66)$

$P_{N_(2) } = 2.766$ bar

We know, total pressure is given by

$P = P_{O_(2) } + P_{N_(2) } + P_{CO_(2) }$

$P_{O_(2) } = 4.50 - 0.269 - 2.776$

$P_{O_(2) } =1.455$ bar

Therefore, the partial pressure of
O_(2) = 1.455 bar.