Question

an escaped helium balloon, initially at a temperature of 21.2 degrees celsius, experiences a pressure drop from from 214000 pa to 201000 pa as it rises into the atmosphere. assuming it maintains its volume of 19.4 litres, by how much (in si units) does the helium temperature change?

Answer 1

In answering the question about how much the helium temperature changes in an escaping balloon at constant volume, it is determined that the temperature decreases by 17.8 Kelvin due to the pressure drop from 214000 Pa to 201000 Pa.

Step-by-step explanation:

The student's question relates to the temperature change of a helium balloon as it experiences a pressure drop while maintaining its volume. This is a physics problem involving the Gas Laws, specifically Gay-Lussac's Law, which relates pressure and temperature of a gas at constant volume. The formula for Gay-Lussac's Law is №₁/ᶜ₁ = №₂/ᶜ₂ where № is the pressure and ᶜ is the temperature in Kelvins. The temperature change Δᶜ can be calculated by rearranging the formula to Δᶜ = ᶜ₂ - ᶜ₁ = №₂/№₁ × ᶜ₁.

Firstly, we convert the initial temperature from Celsius to Kelvin: ᶜ₁ = 21.2°C + 273.15 = 294.35K.

Then, we apply the pressures and initial temperature into the Gay-Lussac's Law formula to find the final temperature ᶜ₂:

ᶜ₂ = (201000 Pa / 214000 Pa) × 294.35K

ᶜ₂ = 0.93925 × 294.35K = 276.55K

Now, to find the temperature change in SI units (Kelvin):

Δᶜ = 276.55K - 294.35K = -17.8K

Therefore, as the helium balloon rises and experiences the pressure drop, its temperature decreases by 17.8 Kelvin.