At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temperature and pressure if you need a total of 1.60 mol of oxygen gas in the vessel

Givens

V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole  * oK
P =

Formula

PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa

You have to add 1.6 – 0.864 = 0.736 moles of gas.

We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.

PV = n*R*T

V = 0.736 * 8.314 * 273 / 490

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41

V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 =

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