Ashampoo bottle contains 553ml of air at 65c what is its volume when it cools to 22c

V₂ = 0.483L = 483mL

Explanation:

Given parameters:

Inital volume of air in shampoo bottle V₁ = 553mL

Original temperature of the bottle T₁ = 65°C

Final temperature T₂= 22°C

Final volume of the gas in the bottle V₂=?

We first take the given units into standard units:

K = T°C + 273

For T₁ we have the temperature to be = 65 + 273 = 338K

                 T₂ = 22 + 273 = 295K

For the volume we convert to Litres or dm³

              1000mL = 1L

V₁ = 553 x 10⁻³L = 0.553L

Now apply one of the gas laws to solve this problem.

From Charles law, we know that the volume of fixed mass of gas is directly proportional to its absolute temperature if the pressure is constant.

Charles's law is expressed as:

The unkown in the equation is V₂ we the express it as the subject of the formula :

                    V₂ =

Putting the parameters in the equation gives:

                   V₂ =

                  V₂ = 0.483L = 483mL