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2 November, 14:19

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm

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  1. 2 November, 14:48
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    85.7°C

    Explanation:

    Step 1:

    Data obtained from the question. This include the following:

    Volume (V) = 73.3mL = 73.3/1000 = 0.0733L

    Mass of l2 = 0.292g

    Pressure (P) = 0.462atm

    Temperature (T) = ... ?

    Step 2:

    Determination of the number of mole present in 0.292g of I2. This is illustrated below:

    Mass of l2 = 0.292g

    Molar Mass of I2 = 127 x 2 = 254g/mol

    Number of mole of I2 = ... ?

    Number of mole = Mass / Molar Mass

    Number of mole of I2 = 0.292/254

    Number of mole of I2 = 1.15*10¯³ mole

    Step 3:

    Determination of the temperature.

    The temperature in the bulb containing the iodine vapor can be obtained by using the ideal gas equation as follow:

    Volume (V) = 0.0733L

    Pressure (P) = 0.462atm

    Number of mole (n) = 1.15*10¯³ mole

    Gas constant (R) = 0.0821 atm. L/Kmol

    Temperature (T) = ... ?

    PV = nRT

    0.462 x 0.0733 = 1.15*10¯³ x 0.0821 x T

    Divide both side by 1.15*10¯³ x 0.0821

    T = (0.462 x 0.0733) / (1.15*10¯³ x 0.0821)

    T = 358.7K

    Converting 358.7K to celsius temperature, we have:

    T (°C) = T (K) - 273

    T (K) = 358.7K

    T (°C) = 358.7 - 273

    T (°C) = 85.7°C

    Therefore, the temperature is 85.7°C
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