Thermodynamic Piston
A cylindrical chamber of cross sectional area 4 c m 2 is filled with an ideal gas, and a piston of mass 2 k g sits at the midpoint of the cylinder. The rest of the chamber has total mass M .
Now, the piston is pulled very slowly upward. If the temperature of the gas inside the chamber remains constant, find the maximum value of M ( in kg ) such that the chamber can be lifted off the ground in this way. Submit your answer to 2 decimal places.
Details and Assumptions:
- There is negligible friction in the system.
- Atmospheric pressure is 1 0 5 N/m 2 , and g = 1 0 m/s 2 .
The answer is 1.00.
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1 solution
I'm struggling to see how the system can be lifted. Lifting the piston simply allows the gas to expand until the piston is removed and the gas fills the container.
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When we lift the container the pressure inside the piston decreases whereas the outside pressure remains the same. If we make the free body diagram of the container (excluding the piston), it will experience two forces, one is gravity in the downward direction and the other is due to the pressure difference.
For the container to lift up before the piston comes out completely, the force of pressure difference must balance the force of gravity on the container.
@Malcom Rich The system will surely lift. See the pressure inside the piston keeps on decreasing and hence the total upward force it will provide at the boundary position will be half of what it was providing when piston was at equilibrium Hence there will be a kind of suction created with upward force + pressure of gas trying to lift piston but are opposed by atm pressure and weight of piston and the latter part will be more since external force does not provide any acceleration hence the piston will lift so that weight of the container comes to play.. The container could have been lifted at any position but I assumed boundary so that I can find the max value of M
What is the fbd for the container just before lifting off, from where did the Mg came to LHS?
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The bottom of the container has air above and below it which exerts a pressure and thus a force on it. When the piston is moved up then the pressure in the container decreases. Due to this, the net force of pressure on the bottom surface of the container is upwards. In addition to it gravity acts on the container in the downward direction. In equilibrium these force are balanced.
Is there a reason to think pulling the piston will make the air expand isothermally?
I guess it might be related to how slowly we move the piston upward, but I'm not practiced enough to recognize these types of features in the problem statement.
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"The piston is pulled upward very slowly." This ensures (1) isothermal expansion and (2) negligible acceleration, so that (3) the gas can be analyzed as being in thermodynamic equilibrium the entire time.
I agree with you, it should be mentioned in the problem that the temperature of the gas remains constant.
If the walls of the container and piston were heat resistant then the process will be adiabatic and it will change the answer.
Let me raise a report for it.
I'm with Malcolm Rich on this. The pressure in the container seems to me to be irrelevant. The overall mass of the container has remained the same and so has it's volume. To lift off, it needs to be less dense than air, but its density has not changed. Alternatively, there needs to be a net force upwards to at least counter gravity, but the pressure in the cylinder is exerted equally in all directions and provides no net force.
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Ah, but the volume (and hence the density) does change. The gas expands into twice its volume before lift-off takes place.
The piston rises because we apply an upward force.
The cylinder rises because the outside air provides an upward force, which is stronger than the downward force from the gas in the container.
The pressure in the cylinder is exerted equally in all directions, but acts on two different things: upward on the piston, and downward on the rest of the cylinder.
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OK thanks, now I see what this problem is about. While the volume of the cylinder which, from the question, is of height h, does not change, the volume of gas does change - previously it occupied only half the cylinder and afterwards it occupies all of it.
In its expanded state, gas has replaced air in the upper half of the cylinder. The cylinder will float if the density of the overall cylinder is less than that of the air it displaces - like a soap bubble. This will require the mass of the air displaced by the cylinder to be at least as great as M+2 kg, assuming the mass of the gas is included in M (or just M if we are continuing to just lift the piston of mass 2kg). This will require a very big cylinder or a very small mass M! I don't believe the calculation at all. Surely it must depend on the value of h.
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@Patrick Ridley – Note that the value of h is far from arbitrary. We're told in the problem that the piston rests at equilibrium at a height of h / 2 .
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@Brian Moehring – Please see my reply to Rohit.
@Patrick Ridley – "This will require the mass of the air displaced by the cylinder to be at least as great as M+2 kg". I think you are forgetting about the force we are applying to lift the piston.
Our force + buoyancy force = total weight of gas + piston + cylinder.
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@Rohit Gupta – You've ignored the rest of the sentence, which is "...(or just M if we are continuing to lift the piston of mass 2kg)". So there are two cases and the question does not make clear which applies, but accepting it means the second, by implication, the force we are applying is only just sufficient to raise the piston and we are applying no force at all to raise the mass of the gas and material of the rest (non-piston) mass of the cylinder. Taken together, these two presumably total M. The pressure of the gas contributes nothing towards raising this mass because it applies equally in all directions - downwards as well as upwards, so that these simply cancel out. Consider Archimedes' principle (https://en.wikipedia.org/wiki/Archimedes%27_principle). Try thinking about the answer given of M = 1 kg. How big a volume of air do you think that would occupy?
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@Patrick Ridley – "The pressure of the gas contributes nothing toward raising this mass..."
Indeed, it does not contribute anything to the net force on the entire system. The force that does the lifting of the system is our upward pull.
But the pressure of the gas does contribute to the "negative pressure" (suction) needed to keep the piston with the rest of the cylinder. More accurately, the outside air pressure pushes piston and cylinder together.
Initially, as we lift the piston, the rest of the cylinder does not move along. We do need more than m g of force, because we pull against the atmospheric pressure; and the pressure inside the cylinder gradually decreases, resulting in the apparent "pull" of a vacuum. If the cylinder were sitting on a scale, the reading on the scale would gradually decrease, until it reaches zero. That is when the cylinder starts moving along with the piston, due to the "suction" of the partial vacuum (or, again, more correctly, due to the outside air pressure doing the lifting for us).
@Patrick Ridley – The force that we will have to apply to lift the piston will be greater than 2 g . This is because, in addition to the weight of the piston, the pressure difference of air below and above the piston will also apply a force on it.
The buoyant force will not be equal to the weight of the cylinder + gas either.
As I said earlier, we can make the following equation. Our force + buoyancy force = total weight of gas + piston + cylinder.
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@Rohit Gupta – OK, thanks, I think I finally understand the meaning of the question. We are supposed to continue to apply the force needed to raise the piston after the piston reaches the top of the cylinder and that is indeed more than 2kg weight because of the increasing downwards pressure difference as the pressure in the cylinder becomes less, in fact becomes half its original value. Thank you for helping me understand the question.
Air pressure (equiv.) is 1Kg per sq cm, or 4Kg on the piston. The additional 2Kg piston mass compresses the gas to 2/3 (i.e. 4/(4+2))of the original (massless piston) volume. So h/2 is 2/3 of the original height, which is therefore 3h/4. Increasing 3h/4 to h (by raising the piston) decreases the pressure in the piston from normal (supporting 4Kg) to 3/4 normal or 3Kg. So the gas in the piston pushes downward on the cylinder bottom with 3Kg while the outside air pushes up with 4Kg. The 1Kg difference is the maximum additional mass M.
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If you read the question correctly then it is stated that initially the system is at equilibrium at H=h/2
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I agree with Aarsh on this, we have to take the initial conditions as given in the question that the piston is at the middle and rests in the equilibrium.
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@Rohit Gupta – This was a question from Fiitjee Aits. I couldnt stop myslef from sharing it...
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@Aarsh Verdhan – That was an interesting problem. Keep sharing such problems. Kudos :)
Since there is no friction, technically chamber will never get the upward force for initial lift-off & it will always remain on the ground. Only when we assume that somehow chamber has already just been lifted off the ground, can we apply atmospheric pressure acting upwards to balance the Mg + Pgas.
@Aarsh Verdhan missed mentioning this important assumption in the solution.
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The cylinder sitting on the ground has a thin layer of air between its base and the ground. The actual area of contact is very tiny. So, the cylinder can still be lifted off the ground.
Actually I cosidered it, but somehow my diagram was changed due to a report. I had no idea about it since I was busy in my exams.
Let the initial pressure of gas be p , then by the ideal gas law p V ∼ T the final pressure will be 2 1 p (for max M ) .
Now from the equilibrium condition p A = m g + P atmosphere A
And just before lifting off 2 1 p A + M g = P atmosphere A
Subsituting given values we get M = 1 k g .