Question

1. choose the the answer.

a balloon was inflated to a volume of 5.00 l at a temperature of 7 °c. it was left in a car on a warm spring day and was heated to 147 °c. what is its new volume?
10.5 l
7.5 l
3.33 l

Explanation:

Step1: Convert temps to Kelvin

$T_1 = 7^\circ\text{C} + 273.15 = 280.15\ \text{K}$
$T_2 = 147^\circ\text{C} + 273.15 = 420.15\ \text{K}$

Step2: Apply Charles's Law

Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Rearrange for $V_2$: $V_2 = V_1 \times \frac{T_2}{T_1}$

Step3: Substitute values

$V_2 = 5.00\ \text{L} \times \frac{420.15\ \text{K}}{280.15\ \text{K}} \approx 5.00\ \text{L} \times 1.5 = 7.5\ \text{L}$

Answer:

7.5 L