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21 December, 08:06

A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increased by 320.4 L at the same pressure?

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  1. 21 December, 08:20
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    T2 = 7.3°C

    Explanation:

    To solve this problem we will use Charles law equation i. e,

    V1/T1 = V2/T2

    Given data

    V1 = 269.7 L

    T1 = 6.12 °C

    V2 = 320.4 L

    T2=?

    Solution:

    Now we will put the values in equation

    269.7 L / 6.12°C = 320.4 L / T2

    T2 = 320.4 L * 6.12°C / 269.7 L

    T2 = 1960.85 °C. L / 269.7 L

    T2 = 7.3°C
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