A balloon was filled to a volume of 2.55 L when the temperature was 30.0∘C. What would the volume become if the temperature dropped to 11.0∘C.
To determine the volume of the balloon when the temperature drops from 30.0∘C to 11.0∘C, we can apply Charles's law, which states that the volume of a gas is directly proportional to its temperature, assuming the pressure and amount of gas remain constant.
Charles's law can be represented in the following formula:
V₁ / T₁ = V₂ / T₂
Where:
V₁: initial volume (2.55 L)
T₁: initial temperature (30.0∘C + 273.15) (Converted to Kelvin by adding 273.15 to Celsius temperature)
V₂: final volume (unknown)
T₂: final temperature (11.0∘C + 273.15) (Converted to Kelvin by adding 273.15 to Celsius temperature)
First, let's convert the temperatures from Celsius to Kelvin:
T₁ = 30.0∘C + 273.15 = 303.15 K
T₂ = 11.0∘C + 273.15 = 284.15 K
Now we can substitute the known values into the formula to find the final volume (V₂):
V₁ / T₁ = V₂ / T₂
2.55 L / 303.15 K = V₂ / 284.15 K
To solve for V₂, we can cross-multiply and then divide:
V₂ = (2.55 L * 284.15 K) / 303.15 K
Calculating this expression:
V₂ ≈ 2.39 L
Therefore, when the temperature drops to 11.0∘C, the volume of the balloon would become approximately 2.39 L.