A hot air balloon starts with its temperature at 68.7°C and a pressure of 0.987 ATM and volume of 564L at what temperature in de

Question

3 years ago

5

grees Celsius while its pressure be 0.852 ATM and its volume be 625L

1 answer:

ICE Princess25 [194] · Answer 1 · 2022-06-10T15:54:52+03:00

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂

1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C