Animated Solution for Physics - Thermodynamics: Comprehension Passage

This problem is a beautiful amalgamation of thermodynamics and fluid mechanics. It tests your ability to visualize physical states, apply the ideal gas law under isothermal conditions, and balance forces using hydrostatics. Let's break down this three-part journey step-by-step.

The Open Hole

Imagine the cylinder with the small hole at the top left completely open. When you slowly pull the piston down to a distance of $2L$, what happens to the air inside? Because the hole is open to the atmosphere, air can freely flow in and out. The system is in constant communication with the outside world.

Therefore, the pressure inside the cylinder simply cannot build up or drop; it remains perfectly equal to the atmospheric pressure, $p_0$. This is a classic conceptual check—don't overthink it when the system is open!

Sealing and Releasing the Piston

Now, the plot thickens. We seal the hole while the piston is at $2L$. At this exact moment, we have trapped a specific amount of air. Its initial pressure is $p_1=p_0$, and its initial volume is $V_1=A(2L)$, where $A=\pi R^2$ is the cross-sectional area of the cylinder.

When we release the piston, it moves to a new equilibrium position, $L^{\prime }$. To find this new position, we must first understand the forces acting on the piston. Let's draw a free body diagram.

Downward, we have the weight of the piston, $Mg$, and the force exerted by the newly compressed trapped air, $pA$. Upward, we have a crucial force: the atmospheric pressure pushing on the open bottom of the cylinder, $p_{0}A$. Equating these forces gives us the new pressure of the trapped air:

Since the cylinder is thermally conducting and the process is slow, the temperature remains constant. We can confidently apply Boyle's Law ($p_1{V}_1=p_2{V}_2$) for this isothermal process:

Substituting our expression for $p$ and solving for $L^{\prime }$:

The Submerged Cylinder

In the final act, we remove the piston entirely, seal the top hole, and plunge the cylinder into a water tank. The water rises inside the cylinder to a height $H$. The trapped air is now compressed into the upper section of length $L_0-H$.

Before submersion, the sealed cylinder was full of air at atmospheric pressure $p_0$, occupying a volume $V_1=A{L}_0$. After submersion, the new volume is $V_2=A(L_0-H)$. Using Boyle's Law again, we find the new pressure $p$ of the trapped air:

Now, we turn to hydrostatics. Consider the horizontal plane at the water surface inside the cylinder. The pressure just above this surface is the trapped air pressure, $p$. According to Pascal's Law, this must equal the pressure at the exact same depth in the surrounding water tank. Since the top of the cylinder is flush with the tank's water surface, the depth is $L_0-H$. The hydrostatic pressure there is $p_0+\rho g(L_0-H)$.

Equating our two expressions for $p$:

To clean this up, we multiply the entire equation by $(L_0-H)$:

Rearranging the terms yields a neat quadratic equation:

And there we have it! A beautiful synthesis of Boyle's Law and hydrostatic equilibrium.