A 9.10-L container holds a mixture of two gases at 29 °C. The partial pressures of gas A and gas B, respectively, are 0.430 atm and 0.558 atm. If 0.150 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Total pressure in initial system = sum of partial pressures
=> Total pressure in initial system = 0.430 atm + 0.558 atm
=> Total pressure in initial system= 0.988 atm
We use the relation PV = nRT to determine the initial moles in the
system
T = (29 + 273) K = 302 K
so, n = PV / RT
= 0.988 atm x 9.10 L / (0.08206 L atm / mol K x 302 K)
= ( 8.9908 / 24.78212) mol
= 0.36279 mol
Total moles in final system = 0.36279 mol + 0.150 mol = 0.51279
mol
New pressure
P = nRT / V
= 0.51279 mol x 0.08206 L atm / mol K x 302 K / 9.10 L
=( 12.70802331/ 9.10 ) atm
= 1.396 atm
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