Water in a container is originally at 100 kPa. The water is subjected to a pressure of 120 MPa. Determine the percentage decrease in its volume.

Percentage decrease in volume = 99.917%

Explanation:

Boyle's Law:States that the volume of a given mass of gas is inversely proportional to the pressure, provided that the temperature is constant.

From Boyle's Law,

P₁V₁ = P₂V₂ ... Equation 1

making V₂ the subject of the equation

V₂ = P₁V₁/P₂ ... Equation 2

Where P₁ = initial pressure, P₂ = final pressure, V₁ = Initial volume, V₂ = final volume

Given: P₁ = 100 kPa = 100*1000 Pa = 100000 Pa, P₂=120 MPa = 120*1000000 Pa = 120000000 Pa

Let: V₁ = y cm³

Substituting these values into equation 2

V₂ = (100000/120000000) y

V₂ = 0.00083y cm³

Percentage decrease in volume = (V₁-V₂/V₁) * 100

Percentage decrease in volume = (y-0.00083y/y) * 100

Percentage decrease in volume = (0.99917y/y) * 100

Percentage decrease in volume = 0.99917*100

Percentage decrease in volume = 99.917%