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 \times Distance moved = 2 cm\(^2\) \times 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 \times 50 = P\(_2\) \times 30 P\(_2\) = \(\frac{(5 \times 50)}{ 30}\) P\(_2\) = \(\frac{250}{30}\) P\(_2\) = \(\frac{25}{3}\) N m\(^{-2}\) (The pressure increases because the volume decreases at constant temperature.)