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15 February, 01:11

Solve this problem using the appropriate law.

A weather balloon with a volume of 50.4 L was launched from a U. S. Navy ship in the Indian Ocean where the temperature was 30 °C and the pressure was 100 kPa. At an altitude of 9000 m the temperature dropped to - 48°C and atmospheric pressure had dropped to 30.8 kPa.

What was the volume of the balloon at that altitude?

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  1. 15 February, 02:04
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    121.5 L will be the new volume

    Explanation:

    To solve this we can use the Ideal Gases Law

    P. V = n. R. T for the two situations. But the moles of gas in the balloon are always the same so:

    P. V / R. T = P. V / R. T

    R is a universal constant, we can reject it. So:

    P₁. V₁ / T₁ = P₂. V₂ / T₂

    We also need to convert T° C to T°K → T°C + 273

    30°C + 273 = 303 K and - 48°C + 273 = 225K

    Let's replace → (50.4L. 100 kPa) / 303K = (30.8 kPa. V₂) / 225K

    ((50.4L. 100 kPa) / 303K). 225K = 30.8 kPa. V₂

    3742.5 kPa. L = 30.8 kPa. V₂

    3742.5 kPa. L / 30.8 kPa = V₂ → 121.5 L
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