Questions: States of Manter Using Charles's Law An arctic weather balloon is filled with 26.1 L of helium gas inside a prep shed. The temperature inside the shed is 6 °C. The balloon is then taken outside, where the temperature is -12 °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.

States of Manter
Using Charles's Law

An arctic weather balloon is filled with 26.1 L of helium gas inside a prep shed. The temperature inside the shed is 6 °C. The balloon is then taken outside, where the temperature is -12 °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.
Transcript text: States of Manter Using Charles's Law An arctic weather balloon is filled with 26.1 L of helium gas inside a prep shed. The temperature inside the shed is $6 .^{\circ} \mathrm{C}$. The balloon is then taken outside, where the temperature is $-12,{ }^{\circ} \mathrm{C}$. 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. $\square$ 1 L $\square$口
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Solution

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Solution Steps

Step 1: Understand Charles's Law

Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) when the pressure is held constant. The formula is given by:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature.

Step 2: Convert Temperatures to Kelvin

To use Charles's Law, we need to convert the temperatures from Celsius to Kelvin. The conversion formula is:

\[ T(K) = T(^{\circ}C) + 273.15 \]

  • Initial temperature \( T_1 = 6^{\circ}C \) becomes \( 6 + 273.15 = 279.15 \, K \).
  • Final temperature \( T_2 = -12^{\circ}C \) becomes \( -12 + 273.15 = 261.15 \, K \).
Step 3: Apply Charles's Law

Using the formula:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Substitute the known values:

\[ \frac{26.1 \, L}{279.15 \, K} = \frac{V_2}{261.15 \, K} \]

Solve for \( V_2 \):

\[ V_2 = \frac{26.1 \, L \times 261.15 \, K}{279.15 \, K} \]

Step 4: Calculate the New Volume

Perform the calculation:

\[ V_2 = \frac{26.1 \times 261.15}{279.15} \approx 24.4 \, L \]

Final Answer

The new volume of the balloon is \(\boxed{24.4 \, L}\).

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