A 8.40-L container holds a mixture of two gases at 11 degrees Celsius. The partial pressures of gas A and gas B, respectively, are 0.163 atm and 0.672 atm. If 0.240 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
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PV = nRT
P = nRT/V = (0.240 mol) x (0.08205746 L atm /K mol) x (11 + 273 K) / (8.40 L) =
0.666 atm, which is the partial pressure of the new gas
So the total pressure becomes: 0.163 + 0.672 + 0.666 = 1.50 atm
P = nRT/V = (0.240 mol) x (0.08205746 L atm /K mol) x (11 + 273 K) / (8.40 L) =
0.666 atm, which is the partial pressure of the new gas
So the total pressure becomes: 0.163 + 0.672 + 0.666 = 1.50 atm