A 9.15-L container holds a mixture of two gases at 37 degrees C. The partial pressures of gas A and gas B, respectively, are 0.443 atm and 0.707 atm. If 0.150 mol a third gas is added with no change in volume or temperature, what will the total pressure become?© BrainMass Inc. brainmass.com December 24, 2021, 10:50 pm ad1c9bdddf
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Calculate the pressure of the third gas by using the ideal gas law.
P is the pressure of the gas.
V is the volume of the gas.
N is the number of moles.
T is the temperature of the gas.
R is the ideal gas constant.
Write down all of the information that is known.
P(A) = Pressure of the gas A = 0.443 atm
P(B) =Pressure of gas B = 0.707 atm
V = 9.15 L
T = 37 C = 310.15 K
R = 0.082 atm L/mol.K
P = nRT/V
Setup the formula and enter the known information for the ideal gas law to determine the pressure of gas C.
P (C)= ((0.150 mol)(0.082 atm-L/mol^-1K)(310.15K))/9.15 L = 0.4169 atm
Calculate the total pressure within the container.
Pt = P(A) + P(B) + P(C) = 0.443 atm + 0.707 atm + 0.4167 atm = 1.5669 atm