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25 March, 18:07

A 7.0-liter sample of N2 (g) at 3.50 atm, a 2.5-liter sample of Ne (g) at 1.00 atm, and a 5.0-liter sample of H2 (g) at 2.45 atm are transferred to a container. The temperature remains constant throughout at 25 °C. The concentration of the mixture in the container is found to be 0.160 mol/L. Calculate the partial pressure of each gas and the total final pressure in the container.

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  1. 25 March, 21:31
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    Total pressure = 6.95 atm

    pN2) 4.34 atm

    p (Ne) = 0.44 atm

    p (H2) = 2.18 atm

    Explanation:

    Step 1: Data given

    The N2 sample = 7.0 liter at 3.50 atm

    The Ne sample = 2.5 liter at 1.00 atm

    The H2 sample = 5.0 L at 2.45 atm

    Temperature = 25.0 °C

    Concentration = 0.160 mol/L

    Step 2: Calculate moles

    p*V = n*R*T

    n = (p*V) / (R*T)

    ⇒ n = the number of moles of gas

    ⇒ V = the volume of gas

    ⇒ p = the pressure of gas

    ⇒ R = the gas constant

    ⇒T = the temperature

    n (N2) = (3.50*7.0) / (0.08206*298 K)

    n (N2) = 1.00 moles

    n (Ne) = (1.00*2.5) / (0.08206*298)

    n (Ne) = 0.102 moles

    n (H2) = (2.45 * 5.0) / (0.08206*298)

    n (H2) = 0.501 moles

    Step 3: Calculate mol fraction

    Total moles = 1.00 + 0.102 + 0.501 = 1.603 moles

    Step 4: Calculate mol fraction

    Mol fraction = moles gas / total moles

    Mol fraction N2 = 1.00 / 1.603 = 0.624

    Mol fraction Ne = 0.102/1.603 = 0.0636

    Mol fraction H2 = 0.501 / 1.603 = 0.313

    Step 5: Calculate total pressure

    Total pressure = 3.50 atm + 1.00 atm + 2.45 atm

    Total pressure = 6.95 atm

    Step 6: Calculate partial pressure

    pN2) = 0.624 * 6.95 = 4.34 atm

    p (Ne) = 0.0636 * 6.95 = 0.44 atm

    p (H2) = 0.313 * 6.95 = 2.18 atm
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