Question

A 9.35 L container holds a mixture of two gases at 17 °C. The partial pressures of gas A and gas B, respectively, are 0.428 atm and 0.628 atm. If 0.190 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

step 1: find the number of mol of in initial mixture before 3rd gas is added

Given:

P = 0.428 atm + 0.628 atm = 1.056 atm

V = 9.35 L

T = 17.0 oC

= (17.0+273) K

= 290 K

find number of moles using:

P * V = n*R*T

1.056 atm * 9.35 L = n * 0.08206 atm.L/mol.K * 290 K

n = 0.4149 mol

step 2: find the final total pressure

Given:

Pi = 1.056 atm

ni = 0.4149 mol

nf = 0.4149 + 0.190 = 0.605 mol

use:

Pi/ni = Pf/Tnf

1.056 atm / 0.4149 mol = Pf / 0.605 mol

Pf = 1.54 atm

Answer: 1.54 atm