Explore BrainMass

Explore BrainMass

    Applying Ideal Gas Law

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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
    https://brainmass.com/chemistry/gas-laws/applying-ideal-gas-law-511487

    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

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:50 pm ad1c9bdddf>
    https://brainmass.com/chemistry/gas-laws/applying-ideal-gas-law-511487

    ADVERTISEMENT