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23 August, 08:06

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

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  1. 23 August, 08:34
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    V = 240.79 L

    Explanation:

    Given dа ta:

    Volume of butane = ?

    Temperature = 293°C

    Pressure = 10.934 Kpa

    Mass of butane = 33.25 g

    Solution:

    Number of moles of butane:

    Number of moles = mass / molar mass

    Number of moles = 33.25 g / 58.12 g/mol

    Number of mole s = 0.57 mol

    Now we will convert the temperature and pressure units.

    293 + 273 = 566 K

    Pressure = 10.934/101 = 0.11 atm

    Volume of butane:

    PV = nRT

    P = Pressure

    V = volume

    n = number of moles

    R = general gas constant = 0.0821 atm. L / mol. K

    T = temperature in kelvin

    V = nRT/P

    V = 0.57 mol * 0.0821 atm. L / mol. K * 566 K / 0.11 atm

    V = 26.49 L/0.11

    V = 240.79 L
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