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22 March, 16:02

What is the pressure exerted by 5.00 moles of nitrogen gas contained in a 30.0 liter container at 25.0 °c?

Answers (2)
  1. S
    22 March, 16:13
    0
    Let's assume that the given gas has an ideal gas behavior. So we can use ideal gas law,

    PV = nRT

    Where P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin (K).

    P = ?

    V = 30.0 L = 30.0 x 10⁻³ m³

    n = 5.00 mol

    R = 8.314 J mol⁻¹ K⁻¹

    T = 25.0 + 273 K = 298 K

    By applying the equation,

    P x 30.0 x 10⁻³ m³ = 5.00 mol x 8.314 J mol⁻¹ K⁻¹ x 298 K

    P = 412928.67 Pa

    P = 4.13 x 10⁵ Pa

    Hence, the pressure of the gas is 4.13 x 10⁵ Pa
  2. V
    22 March, 18:23
    0
    Given: number of moles of N2 = 5

    volume of container = 30 l

    temperature = 25 oC = 298 K

    Also, we know that R = gas consant = 0.082 L atm mol-1 K-1

    Now from ideal gas equation we have,

    PV = nRT

    ∴ P = nRT/V

    = (5 X 0.082 X 298) / 30

    = 4.07 atm.

    Thus, pressure exerted by gas is 4.07 atm
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