At 303 K, the volume of a gas is 30.0 ml. At constant pressure, what is the new volume of the gas if the temperature is decreased to 283K?

V₂ = 28 mL

Explanation:

Given data:

Initial temperature = 303 K

Initial volume = 30.0 mL

Final temperature = 283 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 = 30.0 mL × 283 K /303 K

V₂ = 8490 mL.K / 303 K

V₂ = 28 mL

Kerosene is lighter than water and so it floats on the water and continues to burn. Carbon (IV) oxide blankets the flame cutting off the supply of oxygen and extinguishing the fire.

Explanation:

As water is heavier than kerosene, it slips down permitting the kerosene to rise to the surface and continue to burn. Besides, the existing temperature is so high that the water poured on the fire would evaporate before it can extinguish the fire. Thus the kerosene fire cannot be extinguished with water.