The left and right compartments of a thermally isolated container of length L are separated by a the

Q: Why do we assume the temperatures are equal?

The problem states the piston is 'thermally conducting'. This means heat can freely transfer between the two compartments. In thermodynamic equilibrium, heat transfer stops only when both sides reach the exact same temperature.

Q: How do we know the spring is pulling and not pushing?

The natural length of the spring is $0.4L$, but the piston is at $0.5L$. Since the current length is greater than the natural length, the spring is stretched. A stretched spring always exerts a restoring force that pulls inwards.

Q: Why are the volumes of both compartments equal?

The container has a total length $L$ and a uniform cross-sectional area $A$. The piston is located exactly at a distance of $L/2$ from both edges, dividing the container into two equal halves. Therefore, $V_1 = V_2 = A(L/2)$.

Q: What would happen if the piston was adiabatic instead of thermally conducting?

If the piston were adiabatic, heat could not flow between the compartments. The final temperatures $T_1$ and $T_2$ would depend on the specific thermodynamic process (like adiabatic expansion/compression) and would generally not be equal.

Q: Can we solve this using energy conservation instead of force balance?

While energy is conserved, using it here is unnecessarily complex because we don't know the initial states or the exact path taken to reach equilibrium. Force balance at the final equilibrium state is direct, simple, and gives us exactly what we need.

Question

The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A}\alpha$, then the value of $\alpha$ is _____

Sigma Percentile · Accepted Answer

### Decoding the Physical Setup

Imagine you are looking at a thermally isolated container divided into two compartments by a movable piston.

The problem states a crucial detail: the piston is **thermally conducting**. What does this mean physically? It implies that heat can flow freely between the left and right compartments. In thermodynamic equilibrium, this heat exchange will continue until both sides reach the exact same temperature.

$$T_1 = T_2 = T$$

Furthermore, we are told that the piston rests exactly in the middle of the container, at a distance of $\frac{L}{2}$ from both edges. This geometric constraint guarantees that the volumes of both compartments are identical.

$$V_1 = V_2 = V = \frac{AL}{2}$$

### The Thermodynamic Bridge

Now that we have established the equality of temperatures and volumes, we can apply the Ideal Gas Law to both compartments.

For the left compartment, we have:

$$P_1 V = n_1 R T$$

And for the right compartment:

$$P_2 V = n_2 R T$$

By dividing these two equations, the volume, the universal gas constant, and the temperature beautifully cancel out, leaving us with a direct relationship between the pressures and the number of moles.

$$\frac{P_1}{P_2} = \frac{n_1}{n_2}$$

Substituting the given values of $n_1 = \frac{3}{2}$ and $n_2 = 1$, we find:

$$P_1 = 1.5 P_2$$

### Uncovering the Spring's Secret

Let's shift our focus to the mechanical side of the problem. The left compartment contains a spring attached to the piston.

The natural length of this spring is given as $\frac{2L}{5}$, which is $0.4L$. However, the piston is currently positioned at $\frac{L}{2}$, or $0.5L$.

Because the current length is greater than the natural length, the spring is **stretched**. We can calculate the exact extension $x$ by taking the difference:

$$x = 0.5L - 0.4L = 0.1L$$

### The Mechanical Equilibrium

To find the final pressure, we must analyze the forces acting on the piston. Let's draw a free body diagram.

The gas in the left compartment exerts a force pushing the piston to the right, equal to $P_1 A$. Simultaneously, the gas in the right compartment pushes back to the left with a force of $P_2 A$.

Additionally, because the spring is stretched, it wants to return to its natural state. Therefore, it exerts a restoring force pulling the piston to the left, equal to $kx$.

Since the piston is in mechanical equilibrium, the net force must be zero. The rightward forces must perfectly balance the leftward forces.

$$P_1 A = P_2 A + kx$$

### The Final Calculation

We are now ready to bring everything together. Let's substitute our thermodynamic result ($P_1 = 1.5 P_2$) and our mechanical result ($x = 0.1L$) into the equilibrium equation.

$$1.5 P_2 A = P_2 A + k(0.1L)$$

Subtracting $P_2 A$ from both sides, we isolate the pressure term:

$$0.5 P_2 A = 0.1 kL$$

Solving for $P_2$, we get:

$$P_2 = \frac{0.1 kL}{0.5 A} = 0.2 \frac{kL}{A}$$

The problem states that the pressure in the right compartment is $P = \alpha \frac{kL}{A}$. By comparing this with our derived expression, we can clearly see the final answer.

$$\alpha = 0.20$$

This is a brilliant problem that elegantly weaves together the principles of thermodynamics and Newtonian mechanics!

Enter Numerical Value:

Visualized Solution

$T_{1} = T_{2}, V_{1} = V_{2}$

$P V = n R T$

$\frac{P _{1}}{P _{2}} = 1.5$

$x = 0.5 L - 0.4 L$

$F_{net} = 0$

$P_{1} A = P_{2} A + k x$

$1.5 P_{2} A = P_{2} A + 0.1 k L$

$P_{2} = 0.2 \frac{k L}{A}$

$α = 0.20$

The Sigma Insight: First Law of Thermodynamics

Decoding the Physical Setup

The Thermodynamic Bridge

Uncovering the Spring's Secret

The Mechanical Equilibrium

The Final Calculation

Similar Questions

Comprehension Passage

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be