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11 January, 05:06

A 250 l container with 2.86 kg of propane gas (c3h8) is at a temperature of 30.2 ºc. what is the pressure (atm) of the gas?

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  1. 11 January, 09:04
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    Data Given:

    Pressure = P = ?

    Volume = V = 250 L

    Temperature = T = 30.2 °C + 273 = 303.2 K

    Mass = m = 2.86 Kg = 2860 g

    Moles = n = 2860 g / 44.1 g. mol⁻¹ = 64.85 moles

    Solution:

    Let suppose the gas is acting Ideally, Then According to Ideal Gas Equation.

    P V = n R T

    Solving for P,

    P = n R T / V

    Putting Values,

    P = (64.85 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 303.2 K) : 250 L

    P = 6.45 atm
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