Question

an arctic weather balloon is filled with 14.7 l of helium gas inside a prep shed. the temperature inside the shed is 5. °c. the balloon is then taken outside, where the temperature is 5. °c. calculate the new volume of the balloon. you may assume the pressure on the balloon stays constant at exactly 1 atm. round your answer to 3 significant digits.

Explanation:

Step1: Identify the missing temperature

Wait, the problem seems to have a typo. Let's assume the outside temperature is, for example, -20.0 °C (a common arctic temperature). Let's correct the problem: Initial volume \( V_1 = 14.7 \, \text{L} \), initial temperature \( T_1 = 5.0 \, ^\circ\text{C} = 5.0 + 273.15 = 278.15 \, \text{K} \), final temperature \( T_2 = -20.0 \, ^\circ\text{C} = -20.0 + 273.15 = 253.15 \, \text{K} \) (assuming outside temp is -20°C as arctic). Using Charles's Law (\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) at constant pressure).

Step2: Apply Charles's Law

Rearrange for \( V_2 \): \( V_2 = V_1 \times \frac{T_2}{T_1} \)
Substitute values: \( V_2 = 14.7 \, \text{L} \times \frac{253.15 \, \text{K}}{278.15 \, \text{K}} \)

Step3: Calculate the volume

\( V_2 = 14.7 \times \frac{253.15}{278.15} \approx 14.7 \times 0.9101 \approx 13.38 \, \text{L} \) (rounded to 3 sig figs: 13.4 L)

(Note: If the outside temperature was different, the answer changes. Since the original problem had a typo in the outside temp, we assumed a typical arctic temp. If the actual outside temp is given, recalculate with that \( T_2 \).)

Answer:

Assuming outside temperature is -20.0 °C, the new volume is \(\boldsymbol{13.4 \, \text{L}}\) (adjust if outside temp is different).