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)
This is an application of Boyle’s Law (since temperature is constant):
P\(_1\)V\(_1\) = P\(_2\)V\(_2\)
Given:
- Initial volume, V\(_1\) = 50 cm\(^3\)
- Initial pressure, P\(_1\) = 5 N m\(^{-2}\)
- Distance piston is pushed in = 10 cm
- Cross-sectional area of tube, A = 2 cm\(^2\)
Decrease in volume = Area × Distance moved
= 2 cm\(^2\) × 10 cm
= 20 cm\(^3\)
Calculate the new volume V\(_2\)
V\(_2\) = V\(_1\) − decrease in volume
= 50 cm\(^3\) − 20 cm\(^3\)
= 30 cm\(^3\)
Apply Boyle’s Law
P\(_1\)V\(_1\) = P\(_2\)V\(_2\)
5 × 50 = P\(_2\) × 30
P\(_2\) = \(\frac{(5 × 50)}{ 30}\)
P\(_2\) = \(\frac{250}{30}\)
P\(_2\) = \(\frac{25}{3}\) N m\(^{-2}\)
(The pressure increases because the volume decreases at constant temperature.)
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