50cm\(^3\) of air at a pressure of 5Nm\(^{-2}\) 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 2cm\(^{-2}\)? (Assume that the temperature remains constant throughout)
12.5 Nm-2
8.7 Nm-2
8.3 Nm-2
3.8Nm-2
3.0Nm-2
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Discussions (9)
My School the Answer is C 
First, write down what you know:
Volume air initial, Vi = 50 cm3
Pressure of air initial, Pi = 5 N/m2
Temperature of air = assume constant
If piston is pushed in by 10 cm, the volume decreases by 10 cm • 2 cm2 =
20 cm3
So Volume of air final, Vf = Vi -20 cm3 = 50 cm3 - 20 cm3 = 30 cm3
Since temperature and moles are constant, Boyle’s law applies, which says that pressure and volume are inversely proportional, so if final volume is
30 cm3/50 cm3 = 30/50 of original volume, then final pressure will be the inverse of that, 50/30 multiplied by the original pressure.
5 N/m2 • 50/30 = 8.3 N/m2 or 8 N/m2 (using significant figures)
My School the answer to this question is C and here is prove
To solve this problem, we can use Boyle's Law, which states that:
P1V1 = P2V2
Where:
- P1 is the initial pressure (5 N/m²)
- V1 is the initial volume (50 cm³)
- P2 is the final pressure (unknown)
- V2 is the final volume (unknown)
First, we need to calculate the final volume (V2). The piston is pushed in a distance of 10 cm, which reduces the volume of the air. The cross-sectional area of the tube is 2 cm², so the volume displaced by the piston is:
ΔV = Area x Distance
= 2 cm² x 10 cm
= 20 cm³
The final volume (V2) is the initial volume minus the volume displaced:
V2 = V1 - ΔV
= 50 cm³ - 20 cm³
= 30 cm³
Now, we can use Boyle's Law to find the final pressure (P2):
P1V1 = P2V2
5 N/m² x 50 cm³ = P2 x 30 cm³
P2 = (5 N/m² x 50 cm³) / 30 cm³
= 8.33 N/m²
Therefore, the pressure of the air becomes approximately *8.33 N/m²*.
