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28 January, 16:39

At constant volume, the pressure of a gas is 755 kPa at 30.0 degrees C. What is the temperature of the gas if the pressure is decreased to 252 kPa?

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  1. 28 January, 18:44
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    T₂ = 101.13 K OR

    - 172.02 °C

    Explanation:

    Given dа ta:

    Initial pressure of gas = 755 Kpa

    Initial temperature = 30.0°C

    Final temperature = ?

    Final pressure = 252 Kpa

    Solution;

    Initial temperature = 30.0°C (30+273 = 303 K)

    According to Gay-Lussac Law,

    The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

    Mathematical relationship:

    P₁/T₁ = P₂/T₂

    Now we will put the values in formula:

    755 Kpa / 303 K = 252 Kpa / T₂

    T₂ = 252 Kpa * 303 K / 755 Kpa

    T₂ = 76356 KPa. K / 755 Kpa

    T₂ = 101.13 K

    Kelvin to °C:

    101.13 K - 273.15 = - 172.02 °C
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