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?

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

i believe it is a.) released into

explanation:

if you're asking what comes out, it's steam.