Question

If a balloon with a volume of 10L at 500kpa is heated until the balloon has a temperature of 300°C, 750atm and now has a volume of 100mL. What was the original temperature of the balloon (in °C)

Answer 1

Answer:

103.6°C

Explanation:

We solve this problem, with the Ideal Gases Law. We know that moles of gas are not changed through time.

P . V = n . R . T

So n (number of moles) and R (Ideal Gases constant) are the same, they are cancelled. In the two different states, we can propose:

P₁ . V₁ / T₁ = P₂ . V₂ /T₂

We need to make some conversions:

100 mL . 1 L/ 1000mL = 0.1L

500 kpa . 1atm / 101.3 kPa = 4.93 atm

300°C + 273 = 573 K

We replace data:

4.93 atm . 10L / T₂ = 750 atm . 0.1L / 573K

4.93 atm . 10L = (750 atm . 0.1L / 573K) . T₂

(4.93 atm . 10L) / (750 atm . 0.1L / 573K) → T₂

T₂ = 376.65 K

We convert K to °C →  376.65 K - 273 = 103.6 °C