Question

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 the partial pressure of CO2 is 0.269 bar . Calculate the partial pressure of O2 .

Answer 1

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.