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20 October, 13:59

If 3.11 mol of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 L, what is the temperature of the sample in degrees Celsius?

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  1. 20 October, 16:48
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    Answer: 617.35 °C

    Explanation:

    Using the ideal gas law,

    PV = nRT

    P = Pressure in (atm or pascal)

    V = Volume in (litres or cubic metres)

    n = number of moles.

    R = Gas constant in (0.0821L. atm/mol. K or 8.314J/mol/K)

    T = Temperature in degree Kelvin or Celsius.

    PV = nRT

    P = 2.91atm, V = 78.13L, n = 3.11

    R = 0.0821L. atm/mol. K

    T = ?

    T = 2.91 x 78.13 / 3.11 x 0.0821 = 227.3583 / 0.255331 = 890.5 K

    Converting from Kelvin to Celsius, using: °C = K - 273.15

    = 890.5 - 273.15

    °C = 617.35
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