Question

A 9.60-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.370 atm and 0.645 atm. If 0.220 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer #1

Initial amount of moles:

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Tmperature

R = ideal gas constant

n = PV/(RT)

n = (0.37+0.645) * (9.6) /(0.082 * (27+273))

n = 0.396 moles initially

now,

after addition

final moles = 0.396 + 0.22 = 0.616 moles

find New pressure if no V,T change:

P1/n1 = P2/n2

P2 = P1*N1/(N2)

P2 = (0.37+0.645) *0.396 / 0.616

P2 = 0.6525 atm will be the new pressure, according to ideal gas law ratios

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