Question

In an exhaust pump using a barrel of certain volume, the pressure falls to half of its initial value in just two strokes. The
percent fraction of volume of gas removed with each stroke is approximately​

Answer 1

Answer:

The gas in barrel of pump and vessel is being compressed into volume of vessel in each stroke.

After first stroke,

P 1 V=P 0 (V+v)

⟹P 1 =P 0 ( VV+v )

After second stroke, $

P 2 V=P 1 (V+v)

⟹P 2 =P 0 ( VV+v ) 2

Similarly after n strokes, P n =P 0 ( VV+v ) n

Answer 2

The percent fraction of volume of gas removed with each stroke is approximately 41.4 %

Suppose, the initial pressure = P

And the initial volume is = V

Let the stroke volume be v, i.e., the volume of gas removed per stroke is v.

After first stroke, pressure, P₁ = PV/ (V+v)

After second stroke, pressure, P₂ = P₁V/ (V+v)

                                                        = PV²/(V+v)²

Given,

After two strokes, the pressure falls to half.

So, P₂ = P/2

∴ 1/2 = V²/(V+v)²

⇒ V/(V+v) = 1/√2

⇒ (V+v)/V = √2

⇒ 1 + v/V = √2

⇒ v/V = √2-1

⇒ v/V = 1.414 - 1

⇒ v/V = 0.414

⇒ (v/V × 100) % = 41.4 %

The percent fraction of volume of gas removed with each stroke is approximately 41.4 %