You do a simple experiment at home with the plastic body of a syringe, where you close the exit with one thumb, and push in the plunger with your other thumb. You are able to compress the air inside the syringe from 5ml to 1ml. Assume that you hold the plunger for long enough such that the temperature equalizes to ambient conditions. Required:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?

a. 89.5 N b. 1.59 N

Explanation:

a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?

Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure

So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml

So, making P₂ subject of the formula, we have

P₂ = P₁V₁/V₂

Substituting the values of the variables into the equation, we have

P₂ = P₁V₁/V₂

P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml

P₂ = 1.013 × 10⁵ N/m² × 5

P₂ = 5.065 × 10⁵ N/m²

Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.

So, F = PA

F = P₂πd²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4

F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4

F = 35.8 × 10/4 N

F = 8.95 × 10

F = 89.5 N

b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?

Since the pressure in the syringe after compression is constant, we have

P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of  opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.

So, F₁ = P₂A₁

F = P₂πd₁²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4

F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4

F = 15.91 × 10⁻¹

F = 1.591 N

F ≅ 1.59 N

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