Question

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?

Answer #1

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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