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12 January, 11:25

Calculate the pressure of 3.2 moles of an ideal gas that occupies a volume of 87 m3 at a temperature of 312 K.

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Answers (2)
  1. 12 January, 11:40
    0
    The correct answer is 95.36 Pa.

    Explanation:

    It is given that the moles of an ideal gas, n = 3.2 moles

    Volume of an ideal gas, V = 87 m³

    Temperature, T = 312 K

    Pressure, P = x

    The ideal gas equation, PV = nRT, here R is the gas constant, and at standard temperature and pressure, one mole of ideal gas holds 22.4 L volume, At STP, T = 0 degree C = 273 K

    V = 22.4 L

    moles, n = 1 mole

    Gas constant, R = PV / nT

    R = 1 atm * 22.4 L / 1 mole * 273 K

    1 atm = 1.013 * 10⁵ Pa and 1L = 10⁻³ m³

    R = 8.31 Pa. m³. mol⁻¹. K⁻¹

    Now, there is a need to calculate pressure, P:

    PV = nRT

    P = 3.2 mol * 8.31 Pa. m³. mol⁻¹. K⁻¹ * 312 K / 87 m³

    P = 8296.7 / 87

    P = 95.36 Pa
  2. 12 January, 11:46
    0
    P = 0.0009417 atm

    Or,

    P = 9.417 * 10⁻⁴ atm

    Or,

    P = 0.0954157 kPa

    Or,

    P = 0.715677 mmHg (Torr)

    Explanation:

    Data Given:

    Moles = n = 3.2 mol

    Temperature = T = 312 K

    Pressure = P = ?

    Volume = V = 87 m³ = 87000 L

    Formula Used:

    Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,

    P V = n R T

    where; R = Universal Gas Constant = 0.082057 atm. L. mol⁻¹. K⁻¹

    Solving Equation for P,

    P = n R T / V

    Putting Values,

    P = (3.2 mol * 0.082057 atm. L. mol⁻¹. K⁻¹ * 312 K) : 87000 L

    P = 0.0009417 atm

    Or,

    P = 9.417 * 10⁻⁴ atm

    Or,

    P = 0.0954157 kPa

    Or,

    P = 0.715677 mmHg (Torr)
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