Ask Question
24 February, 23:56

7. What is the total pressure of the following mixture of gases in a 20.0 L container at 298 K: 12.0 g He,

4.00 mol H2 and 6.25 atm Ne?

+2
Answers (1)
  1. 25 February, 00:14
    0
    P (total) = 14.81 atm

    Explanation:

    According to the Dalton law of partial pressure,

    The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.

    P (total) = P1 + P2 + P3 + ... + Pn

    Given dа ta:

    Volume of container = 20.0 L

    Temperature = 298 K

    Mass of He = 12.0 g

    Moles of hydrogen = 4.00 mol

    Pressure of Ne = 6.25 atm

    Solution:

    Moles of helium:

    Number of moles = mass/molar mass

    Number of moles = 12 g / 4 g/mol

    Number of moles = 3 mol

    Pressure of helium:

    PV = nRT

    P = nRT / V

    P = 3 mol * 0.0821 atm. L. mol⁻¹. k⁻¹ * 298 K / 20 L

    P = 73.4 atm / 20

    p = 3.67 atm

    Pressure of hydrogen:

    PV = nRT

    P = nRT / V

    P = 4 mol * 0.0821 atm. L. mol⁻¹. k⁻¹ * 298 K / 20 L

    P = 97.86 atm / 20

    P = 4.89 atm

    Total pressure:

    P (total) = P (He) + P (Ne) + P (H₂)

    P (total) = 3.67 atm + 6.25 atm + 4.89 atm

    P (total) = 14.81 atm
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “7. What is the total pressure of the following mixture of gases in a 20.0 L container at 298 K: 12.0 g He, 4.00 mol H2 and 6.25 atm Ne? ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers