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4 March, 10:34

A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20° C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is - 15° C

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  1. 4 March, 10:59
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    Then volume of the balloon will increase to 7.11 L

    Explanation:

    Step 1:

    Data obtained from the question. This includes:

    Initial pressure (P1) = 1.05 atm

    Initial volume (V1) = 5.0 L

    Initial temperature (T1) = 20°C

    Final pressure (P2) = 0.65 atm

    Final temperature (T2) = - 15°C

    Final volume (V2) = ?

    Step 2:

    Conversion of celsius temperature to Kelvin temperature. This is illustrated below

    K = °C + 273

    T1 = 20°C + 273 = 293K

    T2 = - 15°C + 273 = 258K

    Step 3:

    Determination of the final volume.

    Applying the general gas equation P1V1/T1 = P2V2/T2, the final volume of the balloon can be obtained as follow:

    P1V1/T1 = P2V2/T2

    1.05 x 5/293 = 0.65 x V2/258

    Cross multiply to express in linear form

    293 x 0.65 x V2 = 1.05 x 5 x 258

    Divide both side by 293 x 0.65

    V2 = (1.05 x 5 x 258) / (293 x 0.65)

    V2 = 7.11 L

    Therefore, the volume of the balloon will increase to 7.11 L when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is - 15° C
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