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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.

User Pyjamas
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Answer:

Percentage decrease in volume = 99.917%

Step-by-step 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%

User Aebsubis
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