How does the level of manometer liquid column change as the air is heated on the side of the tank (not exposed to atmosphere)? [closed]

Question

"A cylinder of diameter $d=25.25 cm$, volume $V_i=0.01m^3$ has a piston attached to a spring. A mercury manometer connected to the cylinder is initially as shown in the figure. " The air is now heated steadily until the piston rises slowly a distance of 20 cm. During this process at any instant, the bottom surface of the piston and the mercury meniscus in the open limb are at the same level. Neglecting the weight of the piston and with $\rho_{mercury}=13600 > kg/m^3$,Find final air pressure

Diagram 1:

I see that both volume and temperature increases, meaning that I cannot predict what happens to the pressure at the end of heating.

My approach:

1. At the level of piston bottom, at the end of heating, the pressure is $P_{\text{atm}}$, since the open limb of manometer coincides with this level,
2. Then, $P_{tank}=P_{\text{atm}}+\rho_{air}gH_{tank}$, where $H_{tank}$ can be found from the problem details.

But then, I am faced with the confusion of the behavior of pressure at equilibrium. Is it to that the bottom of the tank, or an average pressure of the tank. In the second case, my approach will not make sense.

Answer 1

The point I am having trouble is: How does the level of manometer liquid column change as the air is heated on the side of the tank (not exposed to atmosphere)? I see that both volume and temperature increases, meaning that I cannot predict what happens to the pressure at the end of heating.

Good! You recognize, for a start, that the manometer height tells you the pressure involved. You can assume that if it's a good manometer then its area is negligible so that it's not impacting the volume much when it rises or falls.

I understand why you would be concerned about all of these parameters, and in fact if you assume an ideal gas then you have all that you need to determine everything about this system, assuming that the manometer is negligibly thin. (The insight there is that if you start from $p, V, T$ and go to $p + \delta p, V + \delta V, T + \delta T$, then the ideal gas formula $pV = n R T$ gives you $p~\delta V + V~\delta p + \delta V~\delta p = n R \delta T.$) Furthermore, the condition that the mercury rises perfectly with the height of the piston gives you the spring constant of the spring on the piston as $k = 2~\rho_\text{Hg} ~g~A$ where $A = \frac\pi 4 d^2$ is the area of the piston and $g\approx 9.81 \text{ N}/\text{kg}$ is the local gravitational acceleration.

But you don't need to do that here. The manometer is a pressure-measurer, and the manometer has these two heights $h_\text{top} + h_\text{bottom} = \text{constant},$ with the measurement $$p = p_\text{atm} + \rho_\text{Hg}~g~(h_\text{top} - h_\text{bottom}).$$ You know that $h_\text{top}$ has increased by 20cm, so to keep the constraint constant, $h_\text{bottom}$ has lowered by 20cm, so $h_\text{top} - h_\text{bottom}$ has changed from 10cm to 10 + 20 + 20 = 50 cm. That tells you $p$. You know the final pressure of air in the cylinder because you're measuring it.

But then, I am faced with the confusion what pressure does the manometer equilibriate to. Is it to that the bottom of the tank, or an average pressure of the tank. In the second case, my approach will not make sense.

The difference between these is something like $\frac 12 \rho_\text{air}~g~V / A$, so if the "air" has anything like its normal density (1 kg/m^3) then we're talking an effect which is 10,000 times weaker, in terms of mmHg, than the actual pressure differences you're measuring. You shouldn't worry about those things.

Answer 2

You should assume the the air pressure inside the tank is uniform once the piston reaches its end state. It won't be exactly, but the differences within the tank will be of trivial magnitude, yet very difficult to calculate.

The air is now heated steadily until the piston rises slowly a distance of 20 cm. During this process at any instant, the bottom surface of the piston and the mercury meniscus in the open limb are at the same level.

If the piston rose 20 cm, then the mercury meniscus in the open limb rose 20cm. That mercury has to come from somewhere, so the meniscus in the tank limb has to have fallen by 20 cm. That means the difference between the two is now $(20+20+10) cm = 0.5m$. This won't last of course - as the air cools, the air inside will contract, reducing pressure. For the steady state instant they describe, however, it is still just as standard manometer problem

($p_{airintank} = p_{atm}+\rho_{mercury} * g * h $) .

Where $h = 0.5m$

EDIT: Significant changes because I misread the question.