To what temperature must a sample of helium
gas be cooled from 123�C to reduce its volume
from 3.7 L to 0.27 L at constant pressure?
Answer in units of K
(V1/T1) = (V2/T2)
To determine the required temperature, we can use the combined gas law equation:
(P₁V₁) / T₁ = (P₂V₂) / T₂
Where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
In this case, the pressure is constant, so the equation simplifies to:
(V₁) / T₁ = (V₂) / T₂
Rearranging the equation to solve for T₂:
T₂ = (V₂ * T₁) / V₁
Now we can plug in the given values:
V₁ = 3.7 L
V₂ = 0.27 L
T₁ = 123 ºC + 273.15 (converting to Kelvin)
T₂ = (0.27 L * (123 ºC + 273.15)) / 3.7 L
Calculating:
T₂ ≈ 81.671 K
Therefore, to reduce the volume of the helium gas from 3.7 L to 0.27 L at a constant pressure, the sample needs to be cooled to approximately 81.671 K.