As shown in the figure, an insulated container is fitted with a thermally conducting but immovable p

Q: Why is $S_1$ at constant volume and $S_2$ at constant pressure?

$S_1$ is trapped between fixed, immovable walls, so its volume cannot change. $S_2$, however, has a freely movable piston that is exposed to the atmosphere. Since the piston is in equilibrium, the pressure inside $S_2$ must always equal the constant atmospheric pressure outside.

Q: Why do we use $C_v$ for $S_1$ and $C_p$ for $S_2$?

The heat capacity depends on the process. For $S_1$ (constant volume), all heat goes into changing internal energy, so we use $C_v$. For $S_2$ (constant pressure), some heat also does work as the piston moves, so we must use $C_p$.

Q: Why is there a negative sign in the differential equation for $\Delta T$?

The negative sign indicates that the temperature difference $\Delta T$ is decreasing over time. As heat flows from the hotter to the colder gas, their temperatures converge, reducing the difference.

Q: What if the gas was diatomic instead of monoatomic?

The values of the molar heat capacities would change. For a diatomic gas, $C_v = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$. This would change the constants in our differential equation, ultimately leading to a different value for $n$.

Q: Does it matter which section is initially hotter?

No, the math works out the same. If $S_2$ was hotter, heat would flow the other way, $dT_2$ would be negative, and $dT_1$ would be positive, but the rate of change of the absolute temperature difference would follow the exact same differential equation.

Question

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition ($P_1$) and a freely movable but thermally insulated piston ($P_2$). The partition $P_1$ with thermal conductivity $K$, cross sectional area $A$ and width $x$ divides the container into two sections, $S_1$ and $S_2$, each containing one mole of a monoatomic gas. The piston $P_2$ moves freely such that the gas in $S_2$ is always at the atmospheric pressure. Initially, the difference between the temperatures of $S_1$ and $S_2$ is $\Delta T_0$. The time it takes for the temperature difference to become $\frac{\Delta T_0}{2}$ is $nxR/KA$, where $R$ is the universal gas constant. The value of $n$ is:
[Given: $\ln 2 \approx 0.7$]

Sigma Percentile · Accepted Answer

### The Tale of Two Chambers

Imagine two rooms side by side, separated by a wall that allows heat to pass through but cannot move. Room 1 ($S_1$) is completely rigid—its walls are fixed, meaning its volume can never change. Room 2 ($S_2$), however, has a flexible wall—a freely movable piston exposed to the outside atmosphere. This means the pressure inside Room 2 will always adjust to match the constant atmospheric pressure outside.

Both rooms contain exactly one mole of a monoatomic gas. Because they are at different temperatures, heat will naturally flow from the hotter room to the colder one through the conducting partition. Our goal is to find out exactly how long it takes for the initial temperature difference, $\Delta T_0$, to drop to half its value.

### The Thermodynamics of Heat Exchange

Let's assume $S_1$ is hotter than $S_2$. Heat will flow from $S_1$ to $S_2$. According to Fourier's law of heat conduction, the rate of this heat transfer is directly proportional to the temperature difference $\Delta T = T_{S1} - T_{S2}$:

$$ \frac{dQ}{dt} = \frac{KA \Delta T}{x} $$

As $S_1$ loses heat, its temperature drops. Because its volume is fixed, this is an **isochoric process**. All the heat lost goes directly into decreasing its internal energy. Using the molar heat capacity at constant volume ($C_v = \frac{3}{2}R$ for a monoatomic gas), we can write:

$$ -dQ = n C_v dT_1 \implies dT_1 = -\frac{2}{3R} dQ $$

Meanwhile, $S_2$ absorbs this exact same heat. But $S_2$ is at constant pressure! As it heats up, it expands, doing work against the atmosphere. This is an **isobaric process**. We must use the molar heat capacity at constant pressure ($C_p = \frac{5}{2}R$):

$$ dQ = n C_p dT_2 \implies dT_2 = \frac{2}{5R} dQ $$

### Forging the Master Equation

We are interested in the temperature difference, $\Delta T$. Let's see how fast this difference is shrinking by taking its derivative with respect to time:

$$ \frac{d(\Delta T)}{dt} = \frac{dT_1}{dt} - \frac{dT_2}{dt} $$

Substituting our expressions for the temperature changes of each individual gas:

$$ \frac{d(\Delta T)}{dt} = -\frac{2}{3R} \frac{dQ}{dt} - \frac{2}{5R} \frac{dQ}{dt} = -\frac{dQ}{dt} \left( \frac{2}{3R} + \frac{2}{5R} ight) $$

Adding the fractions gives us $\frac{16}{15R}$. Now, we substitute our very first equation for the rate of heat conduction, $\frac{dQ}{dt}$:

$$ \frac{d(\Delta T)}{dt} = -\frac{16}{15R} \left( \frac{KA \Delta T}{x} ight) = -\frac{16KA}{15xR} \Delta T $$

Notice the beautiful negative sign! It mathematically guarantees that the temperature difference is decaying over time, exactly as physics dictates.

### The Final Countdown

We now have a classic first-order separable differential equation. We separate the variables and integrate from time $t=0$ (where the difference is $\Delta T_0$) to time $t$ (where the difference is $\frac{\Delta T_0}{2}$):

$$ \int_{\Delta T_0}^{\Delta T_0/2} \frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR} \int_0^t dt $$

Evaluating the integral yields the natural logarithm:

$$ \ln\left(\frac{\Delta T_0/2}{\Delta T_0}ight) = -\frac{16KA}{15xR} t $$

$$ \ln\left(\frac{1}{2}ight) = -\frac{16KA}{15xR} t \implies -\ln 2 = -\frac{16KA}{15xR} t $$

Solving for $t$, we get:

$$ t = \frac{15xR}{16KA} \ln 2 $$

We are given that $\ln 2 \approx 0.7$. Plugging this in:

$$ t = \frac{15 	imes 0.7}{16} \frac{xR}{KA} = \frac{10.5}{16} \frac{xR}{KA} = 0.65625 \frac{xR}{KA} $$

Rounding to two decimal places, we find that $t \approx 0.66 \frac{xR}{KA}$. Comparing this to the given form $t = n \frac{xR}{KA}$, we conclude that **$n = 0.66$**.

Enter Numerical Value:

Visualized Solution

\text{Analyzing the System}

\text{Rate of Heat Transfer}

\text{Temperature Change in } S_1

\text{Temperature Change in } S_2

\text{Rate of Change of } \Delta T

\text{Forming the Differential Equation}

\text{Integrating the Equation}

\text{Calculating } n

The Sigma Insight: First Law of Thermodynamics

The Tale of Two Chambers

The Thermodynamics of Heat Exchange

Forging the Master Equation

The Final Countdown

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Visualized Solution

\text{Analyzing the System}

\text{Rate of Heat Transfer}

\text{Temperature Change in } S_1

\text{Temperature Change in } S_2

\text{Rate of Change of } \Delta T

\text{Forming the Differential Equation}

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The Sigma Insight: First Law of Thermodynamics

The Tale of Two Chambers

The Thermodynamics of Heat Exchange

Forging the Master Equation

The Final Countdown

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