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12 April, 02:26

A balloon contains 2500. mL of Helium at 75.00°C. When the temperature is lowered to

-25.00 °C and the pressure remains unchanged, what is the new volume?

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  1. 12 April, 02:32
    0
    +75°C = 273+75 = 348 K

    -25 °C = 273 - 25=248 K

    now as at constant pressure, V/T = constant

    so

    2500/348 = V/248

    Explanation:

    V = 2500/348 * 248 = 1781.6 mL
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