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24 January, 15:46

What is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

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  1. 24 January, 16:08
    0
    -96.03°C

    Explanation:

    We'll begin by writing out the information provided by the question. This includes:

    Number of mole (n) = 0.645 mole

    Volume (V) = 2.00 L

    Pressure (P) = 4.68 atm

    Temperature (T) = ?

    Recall: that the gas constant = 0.082atm. L/Kmol

    With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

    PV = nRT

    4.68 x 2 = 0.645 x 0.082 x T

    Divide both side 0.645 x 0.082

    T = (4.68 x 2) / (0.645 x 0.082)

    T = 176.97 K

    Now, We can also express the temperature obtained in celsius as shown below:

    Temperature (celsius) = temperature (Kelvin) - 273

    Temperature (celsius) = 176.97 - 273

    Temperature (celsius) = - 96.03°C

    The temperature of the Neon gas is

    -96.03°C
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