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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 .

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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.

User Lester Buck
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