A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 5 kg and a diameter of 12 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas inside the cylinder.

A. 95.6 kPa.

B. 115.36 kg.

Explanation:

A.

P1 = Patm + F1/A

Therefore,

Patm = P1 - (m*g)/(D^2 * π/4)

= 100 - (5 * 9.81)/(0.12^2 * π/4) x 10^-3 kPa

= 100 - 4.336

= 95.6 kPa.

B.

Δm = m2 - m1

= (D^2 * π/(4 * g)) * (P2 - Patm) - m1

= (0.12^2 * π/(4 * 9.81)) * (2*100 - 95.6) x 10^3 - 5

= 115.36 kg.

we need to know the time interval during which the force acts. impulse of force = momentum change