A sealed piston contains 400 mL of air at standard ambient pressure. The piston is compressed to an internal volume of 125 mL while keeping a constant temperature,
What is the new internal pressure (in kPa) of the air?

324.24 kPa

Explanation:

Given that;

Initial pressure P1 = 101325 Pa

V1= 400 ml

P2 = ?

V2= 125mL

From Boyle's law;

P1V1 = P2V2

P2 = P1V1/V2

P2= 101325 × 400/125

P2= 324.24 kPa

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