Question

A 8.15-L container holds a mixture of two gases at 31 °C. The partial pressures of gas A and gas B, respectively, are 0.215 atm and 0.833 atm. If 0.160 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? In atm

Answer 1

total pressure initially = sum of partial pressure of A and B = 0.215 atm + 0.813 atm = 1.028 atm

PV = nRT

P = 1.028 atm
V = 8.15 L
n = ? mol
R = gas constant = 0.082057 Latmmol^-1K^-1
T = temp in Kelvin = 273 +31 = 304 K

1.028 x 8.15 = n x 0.0821 x 304

8.3782 = n x 24.95

n = 8.3782 / 24.95

n = 0.336 moles

there are the moles of gas initially

now new moles = 0.336 + 0.160 = 0.5 moles

again

P = ? atm
V = 8.15 L
n = 0.5 mol
R = 0.082057 Latmmol^-1K^-1
T = 304 K

P x 8.15 = 0.5 x 0.0821 x 304

P = 12.5/8.15

P = 1.5. atm