A 8.45-L container holds a mixture of two gases at 55 °C.?
The partial pressures of gas A and gas B, respectively, are 0.189 atm and 0.591 atm. If 0.110 mol a third gas is added with no change in volume or temperature, what will the total pressure become?
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the total pressure of the inital 2 gases = sum of the partial pressures of the 2 gases
total pressure = 0.189 atm + 0.591 atm = 0.780 atm
now use the ideal gas equation to work out the total number of moles of gas
PV = nRT
P = 0.780 atm
V = 8.45 L
n = ? mol
R = 0.082057 Latmmol^-1K^-1
T = 328.15 K
n = PV / RT
= 0.780 atm x 8.45 L / (0.082057 x 328.15 K)
= 0.24477 mol moles of Gas
Now, you added 0.110 mol of a third gas
total moles = 0.24477 mol + 0.110 mol
= 0.35477 mol
Now, plug the new moles back into the ideal gas equation
P = ? atm
V = 8.45 L
n = 0.35477 mol
R = 0.082057 Latmmol^-1K^-1
T = 328.15 K
P = nRT / V
= 0.35477 mol x 0.082057 x 328.15 K / 8.45 L
= 1.13 atm