Question

A balloon is inflated with helium to a volume of 45L at room temperature (298K). If the balloon is cooled to 200.K, what is the new volume of the balloon?

Answer 1

V₂ = 30.20 L

Step-by-step explanation:

Given data:

Initial volume = 45 L

Initial temperature = 298 K

Final temperature = 200 K

Final volume = ?

Solution;

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 45 L × 200 K / 298 k

V₂ = 9000 L.K / 298 K

V₂ = 30.20 L