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

  • #1
Tomy World
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5
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
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  • #2
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.
 
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  • #3
Tomy World said:
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.
 
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  • #4
Steve4Physics said:
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.
 
  • #5
Tomy World said:
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.

Tomy World said:
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?
 
  • #6
Tomy World said:
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.

Tomy World said:
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?
 
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  • #7
Lnewqban said:
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.
 
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FAQ: How Does a Frictionless Piston Affect Mercury Levels in a Manometer?

What is a manometer?

A manometer is a device used to measure the pressure of a gas or liquid. It typically consists of a U-shaped tube filled with a liquid, such as mercury, and one end of the tube is connected to the gas or liquid whose pressure is to be measured.

How does a manometer work?

A manometer works by balancing the pressure of the gas or liquid against the pressure exerted by a column of liquid (often mercury) in the tube. The difference in the height of the liquid columns on either side of the U-tube indicates the pressure difference.

What causes the mercury levels in a manometer to change?

The mercury levels in a manometer change due to differences in pressure between the gas or liquid being measured and the atmospheric pressure. An increase in the pressure of the gas or liquid will push the mercury down on one side and up on the other, while a decrease will have the opposite effect.

How do you calculate the pressure difference using a manometer?

The pressure difference can be calculated by measuring the difference in height (h) between the two columns of mercury. The pressure difference (ΔP) is given by the formula: ΔP = ρgh, where ρ is the density of the mercury, g is the acceleration due to gravity, and h is the height difference.

What are common sources of error when using a manometer?

Common sources of error when using a manometer include parallax error when reading the mercury levels, temperature changes affecting the density of mercury, and leaks or blockages in the manometer tube that can affect the accuracy of the measurements.

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