How Does a Frictionless Piston Affect Mercury Levels in a Manometer?

[Tomy World]

Homework Statement

The figure shows a piston in a cylinder that is connected to manometer. The atmospheric pressure is 103 kPa. The density of mercury in the manometer is 13,600 kg/m3.
Take g as 10 N/kg.

Describe and explain what will happen to the difference in height between the two mercury levels when more mercury is poured into the manometer.

Relevant Equations

Gas pressure = ρgh + Atmospheric pressure

 

[Tomy World]

There is no more conditions given to this question.
I believed that additional mercury will compress the gas inside cylinder, therefore gas pressure increases. As a result, both mercury levels will increase. And the difference of the mercury levels is higher than 0.2m.

 

[Steve4Physics]

Homework Statement: The figure shows a piston in a cylinder that is connected to manometer. The atmospheric pressure is 103 kPa. The density of mercury in the manometer is 13,600 kg/m3.
Take g as 10 N/kg.

Describe and explain what will happen to the difference in height between the two mercury levels when more mercury is poured into the manometer.

Is that the full question? It seems incomplete.

Presumably the question is about the final equilibrium state.

Assuming the system is initially in equilibrium, the inside of the cylinder is not initially at atmospheric pressure. So there must be a force (not shown in the diagram) holding the piston in position.

When the extra mercury is added, some options are:
a) the piston is moved to a new position to keep the pressure in the cylinder at its initial value;
b) the piston is moved to a new position to keep the volume of trapped air constant;
c) the force on the piston is unchanged;
d) the piston is not moved.

I think you need to know which option is used in order to answer the question.

EDIt. Options a) and c) are equivalent. Options b) and d) are only equivalent if the volume of the pipe is negligible compared with the volume of air in the cyclinder.

 

[Tomy World]

Is that the full question? It seems incomplete.

Presumably the question is about the final equilibrium state.

Assuming the system is initially in equilibrium, the inside of the cylinder is not initially at atmospheric pressure. So there must be a force (not shown in the diagram) holding the piston in position.

When the extra mercury is added, some options are:
a) the piston is moved to a new position to keep the pressure in the cylinder at its initial value;
b) the piston is moved to a new position to keep the volume of trapped air constant;
c) the force on the piston is unchanged;
d) the piston is not moved.

I think you need to know which option is used in order to answer the question.

Exactly. I feel the question is incomplete. When I work on this question, I assume piston is not moved. If so, how would you expect the result? Kindly share.

 

[Steve4Physics]

Exactly. I feel the question is incomplete. When I work on this question, I assume piston is not moved.

I don’t think that’s the most likely option, or the diagram would show a fixed volume container rather than cylinder+piston.

However, it’s still an interesting option to consider.

If so, how would you expect the result? Kindly share.

As you may know from the forum rules, we don’t offer solutions to homework – only guidance.

To understand the fixed piston case, I it might help to consider an extreme case. Imagine the cylinder volume is very small and the right hand side of the manometer is very tall..

As you add more and more mercury to the right side what will happen to the pressure of the trapped gas? And hence what will happen to the height-difference between mercury levels?

 

[Lnewqban]

There is no more conditions given to this question.

In that case, I would use the horizontal position of the cylinder to assume that its weight has no effect in this case.
Most problems like this one show either a tank or a vertical piston with some weight on it.

I believed that additional mercury will compress the gas inside cylinder, therefore gas pressure increases. As a result, both mercury levels will increase. And the difference of the mercury levels is higher than 0.2m.

If my above assumption is correct, then your believe is not.
Please, consider the case in which our piston is free to slide sideways with no appreciable friction against the cylinder (second assumption).
In order for the gas pressure to increase, a force must stop the mentioned sliding movement. Which one would?

 

[haruspex]

consider the case in which our piston is free to slide sideways with no appreciable friction against the cylinder

In that case, there would never be a difference in mercury levels. Something would need to inhibit the sliding, but nothing is shown.
It is possible that this is, or was originally, the second part of a two part problem. In the first part, the piston is pushed in some distance to cause the change in levels. In the second part, the piston is held fixed while mercury is added.
So I agree with @Tomy World's view.