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

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