Question

A balloon was filled to a volume of 2.50 l when the temperature was 30.0∘c. What would the volume become if the temperature dropped to 11.0∘c.

Answer 1

Answer:

2.34 L

Explanation:

Assuming the pressure inside the balloon remains constant, then we can use Charle's law, which states that for a gas kept at constant pressure, the ratio between the volume of the gas and its temperature remainst constant:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1 = 2.50 L$ is the initial volume

$V_2$ is the final volume

$T_1 = 30.0^{\circ}C+273 = 303 K$ is the initial temperature

$T_2 = 11.0^{\circ}C+273 = 284 K$ is the final temperature

Substituting into the equation and solving for V2, we find the final volume:

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(2.50 L)(284 K)}{303 K}=2.34 L$