At what temperature will a balloon have a volume of 6.08 L if the temperature is 41.0 ℃ when its volume is 4.87 L? in kelvin

T₂ = 251.6 K

Explanation:

Given data:

Initial temperature = 41°C

Initial volume = 4.87 L

Final volume = 6.08 L

Final temperature = ?

Solution:

Initial temperature = 41°C (41+273.15 = 314.15 K)

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 = 4.87 L × 314.15 K / 6.08 L

T₂ = 1529.9 L.K / 6.08 L

T₂ = 251.6 K

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa