An arctic weather balloon is filled with 34.1L 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 â5.°C. Calculate the new volume of the balloon. You may assume the pressure on the balloon stays constant at exactly 1atm. Be sure your answer has the correct number of significant digits.

When the balloon is taken outside if the outside temperature is lower than the temperature inside the shed, the volume of the helium gas will decrease according to Charles' law. However, if the temperatureoutside is higher than inside the shed, the volume of the helium gas in the weather balloons will increase.

Note: The question is not complete. Assumptionshas been made about the temperature.

Explanation:

According to Charles' law of gases, the volume of a given mass of gas is directly proportional to its temperature in Kelvin provided its pressure remains constant.

Mathematically, V1T1 = V2T2

Given that the volume of the helium gas inside the weather balloon is 34.1 L at a temperature of 5 °C;

When the balloon is taken outside if the outside temperature is lower than the temperature inside the shed, the volume of the helium gas will decrease according to Charles' law. However, if the temperatureoutside is higher than inside the shed, the volume of the helium gas in the weather balloons will increase.

Assuming, the temperature outside the shed was 6.5 °C:

V1 = 34.1 L, T1 = 5.°C or 5.0 × 10^-10 °C = 273.1500000005

V2 = ?, T2 = 6.5 °C 6.5 × 10^-10 °C = 273.15000000065

V2 = V1T1/T2

V2 = (34.1 × 273.1500000005) / 273.15000000065

V2 = 34.1 L

Due the very small increase in temperature, the volume of the balloon increases only very slightly that is increase that it is negligible.