# Applying Ideal Gas Law

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?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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

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