A 8.35-L container holds a mixture of two gases at 33 °C. The partial pressures of gas A and gas B, respectively, are .152 atm and .110 atm.?
If .110 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 ; Total pressure = 0.152 + 0.110 = 0.262 atm
0.262 atm(8.35 liters) = n(0.08205 liter-atm/mole K)(33 + 273)
n = 0.0871 moles
Total moles after addition of third gas = 0.0871 moles + 0.110 = 0.1971
P(8.35) = 0.1971(0.08205)306)
Pressure = 0.593 atm