An arctic weather balloon is filled with 6.15 L of helium gas inside a prep shed. The temperature inside the shed is \( 9 .{ }^{\circ} \mathrm{C} \). The balloon is then taken outside, where the temperature is \( -28 .{ }^{\circ} \mathrm{C} \). 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. 5.34 L \( \square \times 10^{\square} \)

          An arctic weather balloon is filled with 6.15 L of helium gas inside a prep shed. The temperature inside the shed is \( 9 .{ }^{\circ} \mathrm{C} \). The balloon is then taken outside, where the temperature is \( -28 .{ }^{\circ} \mathrm{C} \). 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.
5.34 L
\( \square \times 10^{\square} \)

An arctic weather balloon is filled with 6.15 L of helium gas inside a prep shed. The temperature inside the shed is 9 .^∘C. The balloon is then taken outside, where the temperature is -28 .^∘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.
5.34 L
□× 10^□

Added by Virginia C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Chareen Guzman

09/02/2023

Step 1

- Inside temperature: \( 9^\circ \text{C} = 9 + 273.15 = 282.15 \, \text{K} \) - Outside temperature: \( -28^\circ \text{C} = -28 + 273.15 = 245.15 \, \text{K} \) Show more…

Show all steps

Thanks for your feedback!

An arctic weather balloon is filled with 6.15 L of helium gas inside a prep shed. The temperature inside the shed is \( 9 .{ }^{\circ} \mathrm{C} \). The balloon is then taken outside, where the temperature is \( -28 .{ }^{\circ} \mathrm{C} \). 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. 5.34 L \( \square \times 10^{\square} \)