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18 April, 15:34

2)

If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 25°

C, what will the pressure of the resulting mixture of gases be?

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Answers (1)
  1. 18 April, 15:44
    0
    Total pressure = 6.55 atm

    Explanation:

    Given dа ta:

    Number of moles of oxygen = 5.0 mol

    Number of moles of nitrogen = 3.0 mol

    Volume of tank = 30.0 L

    Temperature = 25°C (25+273 = 298 K)

    Total pressure of mixture = ?

    Solution:

    Pressure of oxygen:

    PV = nRT

    P = nRT/V

    P = 5.0 mol * 0.0821 atm. L/mol. K * 298 k / 30.0 L

    P = 122.33 atm / 30.0

    P = 4.1 atm

    Pressure of nitrogen:

    PV = nRT

    P = nRT/V

    P = 3.0 mol * 0.0821 atm. L/mol. K * 298 k / 30.0 L

    P = 73.4 atm / 30.0

    P = 2.45 atm

    Total pressure = Partial pressure of oxygen + partial pressure of nitrogen

    Total pressure = 4.1 atm + 2.45 atm

    Total pressure = 6.55 atm
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