A hot air balloon is filled with 1.39×106 L of an ideal gas on a cool morning ( 11 ∘C). The air is heated to 119 ∘C.

Question

What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon?

Answer 1

Use Charles's Law to Solve.

V₁ / T₁ = V₂ / T₂

V₁ = 1.39 x 10⁶ L

T₁ = 11 °C → convert to Kelvin = 11 + 273 = 284 K

V₂ = ?

T₂ = 119 °C → convert to Kelvin = 119 + 273 = 392 K

Rearrange the equation to solve for V₂

V₂ = V₁T₂ / T₁ = (1.39 x 10⁶ L x 392 K) / 284 K = 1.92 x 10⁶ L

Volume and Temperature are directly proportional according to Charles's Law. As the temp increases (11 °C to 119 °C), the volume increases (1.39 x 10⁶ L → 1.92 x 10⁶ L).