# How to determine how deep a piston inside a vessel will move under pressure?

I am revising for the GAMSAT (basically general knowledge across all STEM subjects) and I'd appreciate it if someone could help out with a few physics questions which I am relatively new to, they are questions 8 to 11 inclusive, here.

For Question 8, I went by:

• under constant pressure, doubled area = doubled force and vice versa (area and force are directly proportional)

• under constant force, doubled pressure = halved area (pressure and area have an inverse relationship)

• if the area stays the same, doubled force = doubled pressure (force and pressure are directly proportional).

• in hydraulics, doubling pressure doubles the depth of water/double the depth of say the ocean, the pressure there is also doubled.

My original reasoning (which I now think is faulty but was a coincidence that it got me the right answer) was, when the pressure doubled from 100kPa to 200kPa, at 5N of force, the depth the piston travelled downwards (as if the vessel was the ocean) would also double. Since the question asks about the position of the piston in relation to the bottom of the vessel, this means the distance to the bottom of the vessel is halved from 10cm to 5cm. My answer was correct for Q8.

Using this logic, originally I thought, for question 9, 7.5N is 1.5 * 5, so 1.5 * 5cm is 7.5cm. 7.5 - 5cm from base of vessel is 2.5cm, so my answer is 2.5cm. This answer was also correct according to their answer.

But, I think my reasoning is wrong here. Because for question 10, 7.5 = 1.5 * 5, so 1.5* 200 should be 300kPa. The answer they have is 250kPa. Can someone help me out here?

For question 11, why is the answer 600kPa? Since 500kPa is 5x 100kPa, but then 1N is 1/5 of 5N, should it not be 100kPa (the original) times 5 then divided by 5 to give 100kPa, ie, no change?

Incidentally, the questions says "7.5 N of PRESSURE". Is that even correct? Shouldn't it be 7.5 N of FORCE?

• Such questions are not always welcomed here. In future you might try the chat room Problem Solving Strategies. Mar 8, 2022 at 0:08

In these questions we are told that the mass of gas is fixed. We must assume that the temperature of the gas is also constant while the pressure and volume are varied. Thus we apply Boyle's Law which says that pressure x volume is constant for a fixed mass of gas at constant temperature.

Question 8

We are told that a force of 5N applied to the piston doubles the pressure in the vessel (which we must assume is a cylinder). If the pressure doubles then the total force acting downwards on the piston must have doubled also. So the initial force acting on the piston (due to its own weight plus the force from any 'atmosphere' or fluid pressing down on it) must have been 5N. We need to know this for Questions 9-11.

Doubling the pressure results in half the volume, because of Boyle's Law. So the height of the piston above the base decreases from 10cm to 5cm.

Your answer was correct but your reasoning is a bit mixed up. At one point you seem to be thinking of the "vessel" as a boat with vertical sides floating in the ocean. Also you seem to be saying that if the force on the piston is doubled then it will move down twice as far (towards the base?), which is not correct. However you get the right explanation in the end. It is not the distance the piston moves from its original position which is used in the calculation, but the distance which remains between the piston and the base, because this is proportional to the volume of the gas in the cylinder. If the pressure doubles then the volume is halved so this distance is halved. This correctly expresses the inverse relationship between pressure and volume given by Boyle's Law.

Question 9

Applying a force of 7.5N increases the force on the piston from the initial value of 5N to 12.5N - ie by a factor of 2.5 = 10/4. So the volume of gas in the cylinder would decrease by a factor of 4/10. The height of the piston above the base decreases by the same factor because the diameter is constant - ie from 10cm to 4cm. This is not one of the answers available. The question is faulty.

Your reasoning appears to be that if a 5N force makes the piston move down by 5cm then using the same proportions a 7.5N force should make the piston move down by 7.5cm, leaving it 2.5cm above the base.

You can see this must be wrong by asking yourself what would happen if you applied a force of 10N instead of 5N. Would this make the piston go down 10cm? Then it would be pressed flat against the base and the gas inside would be squeezed into a space of 0 volume! This does not happen in practice. No matter how hard you squeeze it, the gas inside the cylinder always occupies a finite volume, however small. It never becomes 0.

You made 2 mistakes. First you did not apply Boyle's Law as you did for Question 8. Second you missed the fact that there must be a force already pressing down on the piston, otherwise the gas in the cylinder would keep expanding until the pressure inside it is zero. (The person who set the question appears to have made this second mistake also.)

Question 10

The force on the piston has increased by a factor of 2.5 (as calculated in Question 9). So the pressure increases by the same factor, because the area of the piston is the same. The answer should be 2.5 x 100kPa = 250kPa.

The mistake you made here is that you did not take into account the initial force of 5N which acts on the piston before any force is applied to it.

The pressure in the cylinder is 100kPa initially, when the total force on the piston (due to its weight plus atmospheric pressure) is 5N. After an applied force of 5N is added the total force becomes 10N. The pressure is then 200kPa - because double the force, double the pressure. If the applied force is 7.5N instead of 5N then the total force on the piston is now 12.5N which is 2.5x the initial force of 5N. So the pressure is 2.5 x 100kPa = 250kPa.

Question 11

This question is confusing and badly constructed.

If the piston and cylinder are the same as before then the total force on the piston without any additional load is still 5N. And it has the same area so the pressure it exerts on the gas (which equals the pressure the gas exerts on it) should again be 100kPa. This will be the same regardless of how much gas is in the cylinder, or what kind of gas it is, or what its temperature is.

However the pressure in the cylinder has somehow increased to 500kPa. How has this happened? We can only guess that an additional force has been applied to the piston, increasing the total force it applies to the gas from 5N (when the pressure was 100kPa) to 25N (when the pressure will be 500kPa).

Now we add a 1N load to the piston making the total force on it increase from 25N to 26N. So the pressure increases in proportion from 500kPa to 26/25 x 500 = 520kPa.

This is not one of the options available. The question is again faulty.

Your reasoning is very wrong. If you start with a pressure of 500kPa and add more force to the piston, the pressure in the gas will increase not decrease.

You seem to have gotten confused by the question, thinking that the total force on the piston has decreased from 5N to 1N. If that had been the case then your answer would be correct. The total force has increased but the question is faulty anyway so it is not surprising you got the wrong answer.

Force v Pressure in Question 10

You are correct. Force is measured in N. Pressure is measured in kPa. 7.5N is a force not a pressure.

• Thank you for this comprehensive answer! To address a few issues brought up: Q8) Yes, I initially thought of it as a boat being pushed downwards towards the bottom of the ocean, so that, it is not the distance (the base of the boat/piston) moved but how far it is from the bottom. -1/3 Mar 8, 2022 at 0:16
• Q9) Actually I was thinking about this set of questions yesterday and I have a new thought. Since it is airtight, wouldn't there be an equilibrium point when, the more pressure put on the piston, the more pressure on the gas in the space underneath and the more pressure the gas can exert, forcing the piston back up, preventing the piston from moving? So how can we tell if the formula used will continue to give reliable answers in terms of how far the piston will move? And, just to be clear, it if moves down 4/10 = 40% of 10, would that not be moving down 4cm leaving 6cm from the bottom? -2/3 Mar 8, 2022 at 0:16
• Q11) Question did not say that 5N is the total atmospheric pressure and the its own weight. Perhaps these are just very bad questions and I should look for better revision sources, however: in this case, do we somehow know/assume that the 5N first mentioned, is its own weight and the atmospheric pressure? With the way the wording is now, I just assumed that 1N is the total force, not an additional 1N. Incidentally, if it were the total, would my reasoning be correct, if 5N produces 500kPa then 1N would produce 100kPa? Thank you in advance Mar 8, 2022 at 0:21
• Q9) When you add a weight suddenly to the piston it oscillates up and down like adding a weight to a spring. The oscillations continue until the kinetic energy of the moving mass or piston is dissipated. Then each reaches an equilibrium position in which the forces on them are balanced and motion stops. Mar 8, 2022 at 1:19
• Q9) The formula (Boyle's Law) gives the correct answer for the final resting position of the piston. It does not tell you the amplitude of oscillation. Like Hooke's Law for springs, Boyle's Law is only an approximation but it is highly accurate for most gases provided that their density is much less than the density when it condenses into a liquid. Mar 8, 2022 at 1:20