A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into

Q: Why is the final temperature of both compartments the same?

The partition is described as 'diathermic', which means it is a perfect thermal conductor. Heat will flow between the top and bottom compartments until thermal equilibrium is reached, ensuring they share the same final temperature.

Q: Why does the area $A$ cancel out in the force equation?

Pressure is force per unit area ($P = F/A$). When we multiply the pressure by the cross-sectional area $A$ to get the force, the $A$ in the denominator of the volume ($V = A \times h$) perfectly cancels out with the $A$ in the numerator.

Q: Why can't we just assume the gas is monatomic?

In JEE Advanced, you should never assume data that isn't explicitly given unless it's a standard constant. The exact final position depends heavily on whether the gas is monatomic, diatomic, or polyatomic. This ambiguity is why the question was declared a bonus.

Q: Why did the examiner likely assume the process was isothermal?

Look at the numbers! $nR = 0.83$ and $mg = 83$. If $T_f = 300$, the product $nRT_f = 249$, which is exactly $3 \times 83$. The math works out so perfectly to a clean quadratic equation that it's almost certain the setter designed it this way, forgetting that the temperature must actually rise.

Q: Where does the lost potential energy of the partition go?

Since the entire cylinder is thermally isolated, no heat can escape to the surroundings. The lost gravitational potential energy of the falling partition is entirely converted into the internal energy of the gas, raising its temperature.

Question

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be ______. (take the acceleration due to gravity = 10 ms^{-2} and the universal gas constant = 8.3 J mol^{-1} K^{-1}).

Sigma Percentile · Accepted Answer

### The Setup: A Deceptive Cylinder

Imagine you are standing in front of a massive, 8-meter-tall cylinder. It is perfectly thermally isolated from the outside world. Inside, a heavy 8.3 kg partition splits the cylinder perfectly in half.

Both the top and bottom compartments hold exactly 0.1 moles of an ideal gas at a comfortable 300 K.

Suddenly, the partition is released! Gravity takes over, pulling the heavy partition downward. It compresses the gas in the lower compartment while allowing the gas in the upper compartment to expand.

Eventually, the system settles into a new equilibrium state. Our mission is to find the new depth of this partition, let's call it $y$.

### The Physics of Equilibrium

When the partition finally comes to rest, it must be in perfect mechanical equilibrium.

This means the upward force exerted by the pressure of the bottom gas ($P_2 A$) must exactly balance the downward forces. These downward forces consist of the pressure from the top gas ($P_1 A$) and the gravitational weight of the partition itself ($mg$).

Mathematically, we can write this as:
$$P_2 A = P_1 A + mg$$

Dividing the entire equation by the cross-sectional area $A$, we get a clean relationship between the pressures:
$$P_2 - P_1 = \frac{mg}{A}$$

### The Temperature Conundrum

The partition is described as "diathermic," meaning it is a perfect conductor of heat.

Because heat can flow freely between the top and bottom compartments, they will eventually reach thermal equilibrium and share the exact same final temperature, $T_f$.

Using the ideal gas law, $P = \frac{nRT}{V}$, we can express the new pressures in terms of this final temperature and the new volumes. The top volume is $Ay$ and the bottom volume is $A(8-y)$.

Substituting these into our force balance equation gives:
$$\frac{nRT_f}{A(8-y)} - \frac{nRT_f}{Ay} = \frac{mg}{A}$$

Notice how beautifully the area $A$ cancels out from every term! We are left with:
$$nRT_f \left( \frac{1}{8-y} - \frac{1}{y} ight) = mg$$

### The Missing Link: Energy Conservation

Here is where the plot thickens and the mystery begins.

The entire cylinder is thermally isolated, meaning no heat enters or leaves the system ($Q = 0$). According to the **First Law of Thermodynamics**, the loss in gravitational potential energy of the falling partition must be entirely converted into the internal energy of the gas.

The partition falls from a depth of 4 meters to a new depth $y$, so it loses $mg(y - 4)$ of potential energy. The gas gains this energy, increasing its temperature from $T_0$ to $T_f$.

We can write this energy conservation equation as:
$$mg(y - 4) = 2nC_v(T_f - T_0)$$

Look closely at this equation. To find the final temperature $T_f$, we desperately need the molar heat capacity at constant volume, $C_v$.

But the problem simply states we have an "ideal gas." It never specifies whether the gas is monatomic, diatomic, or polyatomic! Without knowing the atomicity of the gas, $C_v$ is unknown, and the exact final temperature cannot be calculated.

### The Examiner's Trap

This missing information is exactly why this question was awarded bonus marks in JEE Advanced 2020.

However, let's put ourselves in the shoes of the examiner. What if they made a conceptual error and mistakenly assumed the process was isothermal? What if they assumed the temperature remained constant at 300 K?

Let's test this hypothesis by plugging $T_f = 300 	ext{ K}$ into our force balance equation.

We know $nR = 0.1 	imes 8.3 = 0.83 	ext{ J/K}$, and the weight $mg = 8.3 	imes 10 = 83 	ext{ N}$.

Substituting these values yields:
$$0.83 	imes 300 \left( \frac{1}{8-y} - \frac{1}{y} ight) = 83$$

### The Beautiful Cancellation

Look at how perfectly these numbers are designed!

The product of $0.83$ and $300$ is exactly $249$. When we divide $249$ by $83$, we get exactly $3$. This is no coincidence; the numbers were hand-picked for this exact cancellation.

$$3 \left( \frac{y - (8-y)}{y(8-y)} ight) = 1$$

Simplifying the numerator gives $2y - 8$. Cross-multiplying the denominator gives:
$$3(2y - 8) = 8y - y^2$$

Expanding this out, we get:
$$6y - 24 = 8y - y^2$$

Rearranging all terms to one side reveals a beautifully simple quadratic equation:
$$y^2 - 2y - 24 = 0$$

### The Final Verdict

Factoring this quadratic equation is straightforward:
$$(y - 6)(y + 4) = 0$$

Since the depth $y$ must be a positive physical distance, we discard the negative root. This leaves us with:
$$y = 6 	ext{ m}$$

This was undeniably the intended answer! The examiner crafted the numbers perfectly for an isothermal assumption.

However, physics is unforgiving. In reality, the temperature would absolutely increase due to the dissipated potential energy, making the true depth slightly different and heavily dependent on the specific type of gas.

A brilliant trap, a fascinating physical scenario, and a well-deserved bonus for the students!

Visualized Solution

Initial State Analysis

Releasing the Partition

Mechanical Equilibrium

Pressure Equations

The Force Balance Equation

The Missing Link (Energy Conservation)

The Setter's Trap

Solving the Isothermal Case

Final Calculation

The Intended Answer

The Sigma Insight: First Law of Thermodynamics

The Setup

The Physics of Equilibrium

The Temperature Conundrum

The Missing Link

The Examiner's Trap

The Beautiful Cancellation

The Final Verdict

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Visualized Solution

Initial State Analysis

Releasing the Partition

Mechanical Equilibrium

Pressure Equations

The Force Balance Equation

The Missing Link (Energy Conservation)

The Setter's Trap

Solving the Isothermal Case

Final Calculation

The Intended Answer

The Sigma Insight: First Law of Thermodynamics

The Setup

The Physics of Equilibrium

The Temperature Conundrum

The Missing Link

The Examiner's Trap

The Beautiful Cancellation

The Final Verdict

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