Question

A weather balloon is filled with helium to a volume of 250 L at a temperature of 27°C. The balloon rises to an altitude where the temperature drops to - 33 deg * C at constant pressure. What is the new volume ?

• A. 305 L
• B. 200 L
• C. 255 L
• D. 180 L

Answer 1

Answer:

B. 200 L

• This problem involves the application of Charles's Law, a fundamental principle in the study of gases. Charles's Law describes the relationship between the volume and temperature of a gas when the pressure and the amount of gas remain constant.

Charles's Law Explained

• Charles's Law states that for a fixed amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. Mathematically, this is represented as V₁/T₁ = V₂/T₂.

• It's crucial to use the absolute temperature scale (Kelvin) when applying gas laws. To convert Celsius to Kelvin, add 273.15 (often approximated as 273 for simplicity in these types of problems).

Applying Charles's Law to the Weather Balloon Scenario

• Initial Conditions:

  • Initial Volume (V₁): 250 L

  • Initial Temperature (T₁): 27°C. Converting to Kelvin: 27 + 273 = 300 K

• Final Conditions:

  • Final Temperature (T₂): -33°C. Converting to Kelvin: -33 + 273 = 240 K

  • The pressure is stated to be constant, which is a key condition for Charles's Law.

• Calculation:

  • Using the formula V₁/T₁ = V₂/T₂, we can rearrange to solve for the final volume (V₂): V₂ = V₁ * (T₂/T₁)

  • Plugging in the values: V₂ = 250 L * (240 K / 300 K)

  • V₂ = 250 L * 0.8

  • V₂ = 200 L