Question

A helium balloon in a closed car occupies a volume of 1.32 L at 30.0°C. If the car is parked on a hot day and the volume of the balloon increases to 2.50 L what is the temperature in the car, assuming the pressure remains constant?

Answer 1

T₂ = 573.86 K

Step-by-step explanation:

Given data:

Initial volume = 1.32 L

Initial temperature = 30.0°C (30+273 = 303 K)

Final volume = 2.50 L

Final temperature in car = ?

Solution:

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

T₂ = V₂T₁/V₁

T₂ = 2.50 L × 303 K / 1.32 L

T₂ = 757.5 L.K / 1.32 L

T₂ = 573.86 K