Fig. 4.1 shows a large syringe with the nozzle sealed by wax. A piston is fitted inside the syringe. The air in the syringe is at atmospheric pressure, $1.0 \times 10^5\,$Pa. The volume of the air in the syringe is $1.2 \times 10^{-4}\,$m$^3$. The area of the piston end that is in contact with the air is $3.5 \times 10^{-3}\,$m$^2$. Friction between the piston and the syringe is negligible.
(a(i))[2]
Calculate the force on the piston due to the pressure of the air inside the syringe.
(a(ii))[2]
The force on the piston in (a)(i) acts to the right. Explain why the piston does not move to the right.
(b(i))[3]
The piston is now pulled to the right by an additional force. The temperature of the air in the syringe does not change. Explain, in terms of particles, why the pressure of the air in the syringe decreases.
(b(ii))[2]
Calculate the pressure of the air inside the syringe when the air volume is $1.5 \times 10^{-4}\,$m$^3$.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “F = pA, or $1.0 \times 10^5 \times 3.5 \times 10^{-3}$” …