To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is held constant. Charles's Law can be expressed as:

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

Where:

• $V_1$ is the initial volume of the gas,
• $T_1$ is the initial temperature in Kelvin,
• $V_2$ is the final volume of the gas,
• $T_2$ is the final temperature in Kelvin.

Given:

• $V_1 = 35.3 \, \text{L}$,
• $T_1 = 13^\circ \text{C} = 13 + 273.15 = 286.15 \, \text{K}$,
• $T_2 = -22^\circ \text{C} = -22 + 273.15 = 251.15 \, \text{K}$.

We are tasked with finding $V_2$, the new volume.

Rearrange Charles's Law to solve for $V_2$:

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

Substitute the known values:

$V_2 = 35.3 \, \text{L} \times \frac{251.15 \, \text{K}}{286.15 \, \text{K}}$

Now calculate:

$V_2 \approx 35.3 \, \text{L} \times 0.877$

$V_2 \approx 30.96 \, \text{L}$

Significant Figures:

Since the given volume (35.3 L) has 3 significant digits, the final answer should also be expressed with 3 significant digits.

Thus, the new volume of the balloon is approximately:

$V_2 = 31.0 \, \text{L}$

Would you like further details or have any questions about the calculation?

Here are 5 related questions to explore further:

1. What is Boyle's Law and how does it differ from Charles's Law?
2. How would the calculation change if the pressure were not constant?
3. What are other real-life applications of Charles's Law?
4. How do you convert Celsius to Kelvin in gas law problems?
5. How would the volume change if the outside temperature dropped even further?

Tip: Always convert temperatures to Kelvin when dealing with gas law equations since the Kelvin scale starts at absolute zero, ensuring accuracy in proportionality calculations.