Question

A pump contains 0.5 L of air at 203 kPa . You draw back on the piston of the pump, expanding the volume until the pressure reads 50.8 kPa . What is the new volume of air in the pump?

Answer 1

Answer:

V₂ = 2 L

Explanation:

given,

P₁ = 203 kPa

V₁ = 0.5 L

Now, volume is expanded

P₂ = 50.8 kPa

V₂ = ?

Using Boyle's law

 P₁V₁ = P₂ V₂

$V_2 = \dfrac{P_1V_1}{P_2}$

inserting all the values

$V_2 = \dfrac{203\times 0.5}{50.8}$

     V₂ = 2 L

Hence, new volume of air in the pump is equal to 2 L.