NECO 2001 Physics · Question 6

50cm3 of air at a pressure of 5 Nm^ -2 is enclosed in a metal tube by a piston. If the piston is pushed in a…

Question

50cm3 of air at a pressure of 5 $Nm^{-2}$ is enclosed in a metal tube by a piston. If the piston is pushed in a distance of 10cm, what will the pressure of the air become, if the cross-sectional area of the tube is 2 $cm^{-2}$ ? (Assume that the temperature remains constant throughout)

A.12.5 $Nm^{-2}$
B.8.7 $Nm^{-2}$
C.8.3 $Nm^{-2}$
D.3.8 $Nm^{-2}$
E.3.0 $Nm^{-2}$

Answer: C. 8.3 $Nm^{-2}$

Explanation written with AI and checked against the answer key

Concept: The temperature is constant, so this is Boyle's Law: P₁V₁ = P₂V₂. Step 1: Find the change in volume The piston moves in 10 cm, and the cross-sectional area is 2 cm² (the "cm⁻²" in the question is a typo for cm²). Volume decrease = area × distance = 2 × 10 = 20 cm³ Step 2: Find the new volume V₂ = 50 − 20 = 30 cm³ Step 3: Apply Boyle's Law P₁V₁ = P₂V₂ 5 × 50 = P₂ × 30 P₂ = 250 / 30 ≈ 8.33 Nm⁻² Answer: The pressure becomes about 8.3 Nm⁻², which is option C. Tip: In Boyle's Law questions, first work out the new volume. If the piston is pushed in, the volume decreases and the pressure increases, so your answer must be larger than the starting pressure. Here 8.3 > 5, which confirms it is sensible. This also rules out options D and E, which are smaller than 5.

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50cm3 of air at a pressure of 5 Nm^ -2 is enclosed in a metal tube by a piston. If the piston is pushed in a…

Question

Answer: C. 8.3 Nm−2Nm^{-2}Nm−2

Answer: C. 8.3 $Nm^{-2}$