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24 September, 00:52

A gas‑filled weather balloon with a volume of 59.0 L is held at ground level, where the atmospheric pressure is 752 mmHg and the temperature is 24.3 ∘ C. The balloon is released and rises to an altitude where the pressure is 0.0708 bar and the temperature is - 5.41 ∘ C. Calculate the weather balloon's volume at the higher altitude.

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  1. 24 September, 01:06
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    At the higher altitude, the new volume is 750.2L

    Explanation:

    If we decompose the Ideal Gases Law for the two situations (initial and final), we can determine this relaion:

    P₁. V₁ / T₁ = P₂. V₂ / T₂

    Number of moles does not change, and R stays the same.

    Let's make some conversions, before:

    752 mmHg. 1 atm / 760mmHg = 0.989 atm

    24.3°C + 273 = 297.3K

    0.0708 bar. 0.986 atm / 1bar = 0.070 atm

    -5.41°C + 273 = 267.59 K

    We replace: (0.989 atm. 59L) / 297.3K = (0.070 atm. V₂) / 267.59K

    [ (0.989 atm. 59L) / 297.3K]. 267.59K = 0.070 atm. V₂

    52.5 atm. L = 0.070 atm. V₂

    V₂ = 52.5 atm. L / 0.070atm = 750.2L
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