The rectangular box shown in figure has a partition which can slide without friction along the lengt

Q: Why can't we use the adiabatic equation for the left chamber?

The left chamber has an electric heater inside it, which supplies heat to the gas. Since $\Delta Q \neq 0$, the process is not adiabatic.

Q: Why is the final pressure the same in both chambers?

The partition is free to slide without friction. It will move until the mechanical forces on both sides balance out, which means the pressure must be equal in both chambers at equilibrium.

Q: How did we know the total volume is $2V_0$?

Initially, both chambers have the same pressure $p_0$, temperature $T_0$, and 1 mole of gas. By the ideal gas law, they must have the same initial volume $V_0$. The total volume of the rigid box is $V_0 + V_0 = 2V_0$.

Q: Why is the work done in the right chamber negative?

The gas in the right chamber is being compressed by the expanding gas in the left chamber. Since its volume decreases, the work done BY the gas is negative (work is done ON the gas).

Question

The rectangular box shown in figure has a partition which can slide without friction along the length of the box.
Initially each of the two chambers of the box has one mole of a monoatomic ideal gas ($\gamma = 5/3$) at a pressure $p_0$, volume $V_0$ and temperature $T_0$. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes $243 p_0 / 32$. Determine (a) the final temperature of the gas in each chamber and (b) the work done by the gas in the right chamber.

Sigma Percentile · Accepted Answer

Imagine you are standing in front of a rigid, thermally insulated box. Inside, a frictionless partition divides the box into two identical chambers. Both chambers are perfectly balanced, holding one mole of a monoatomic gas at pressure $p_0$, volume $V_0$, and temperature $T_0$. It’s a state of perfect equilibrium.

But then, we introduce chaos. We turn on an electric heater in the left chamber. As the gas in the left chamber heats up, it expands, pushing the partition to the right. The right chamber, with nowhere to go and no heat escaping, is forced to compress. The partition finally settles when the pressure in both chambers reaches a staggering $\frac{243}{32} p_0$. Our mission? To find the final temperatures of both chambers and the work done by the gas in the right chamber. Let's dive in!

### The Right Chamber: Adiabatic Compression

Let's focus on the right chamber first. It is completely thermally insulated, meaning no heat can enter or leave ($\Delta Q = 0$). As the partition pushes into it, the gas undergoes an **adiabatic compression**.

For an adiabatic process, the relationship between temperature and pressure is governed by the equation:
$$T^\gamma p^{1-\gamma} = 	ext{constant}$$
Which can be rewritten as:
$$T \propto p^{\frac{\gamma-1}{\gamma}}$$

Since the gas is monoatomic, we know $\gamma = \frac{5}{3}$. This makes our exponent $\frac{\gamma-1}{\gamma} = \frac{2}{5}$. Let's set up our equation to find the final temperature $T_2$:
$$T_2 = T_0 \left( \frac{p_f}{p_0} ight)^{2/5}$$

Now, we substitute the final pressure $p_f = \frac{243}{32} p_0$:
$$T_2 = T_0 \left( \frac{243}{32} ight)^{2/5}$$

Here is where the math gets beautiful. Notice that $243 = 3^5$ and $32 = 2^5$. The fifth roots cancel out perfectly!
$$T_2 = T_0 \left( \frac{3}{2} ight)^2 = \frac{9}{4} T_0 = 2.25 T_0$$
The right chamber has heated up significantly just from being compressed!

### The Left Chamber: The Volume Constraint

Now, what about the left chamber? We don't know how much heat the heater supplied, so we can't use the First Law of Thermodynamics directly. But we have a secret weapon: **the total volume of the box is constant**.

Initially, both chambers had a volume $V_0$, making the total volume $2V_0$. No matter where the partition moves, the sum of the final volumes must equal the total volume:
$$V_1 + V_2 = 2V_0$$

Let's find $V_2$ using the ideal gas law ($pV = nRT$):
$$V_2 = \frac{n R T_2}{p_f} = \frac{1 \cdot R \cdot (2.25 T_0)}{\frac{243}{32} p_0}$$
Since $V_0 = \frac{R T_0}{p_0}$, we can simplify this to:
$$V_2 = \frac{9/4}{243/32} V_0 = \frac{8}{27} V_0$$

Now, we can easily find $V_1$:
$$V_1 = 2V_0 - V_2 = 2V_0 - \frac{8}{27} V_0 = \frac{46}{27} V_0$$

With $V_1$ in hand, we apply the ideal gas law one last time to find $T_1$:
$$T_1 = \frac{p_f V_1}{n R} = \frac{\left(\frac{243}{32} p_0ight) \left(\frac{46}{27} V_0ight)}{1 \cdot R}$$
$$T_1 = \frac{9 	imes 46}{32} T_0 = 12.94 T_0$$
The left chamber has reached a blazing $12.94 T_0$!

### The Final Act: Work Done

Finally, we need to calculate the work done by the gas in the right chamber. For an adiabatic process, the work done is simply the negative change in internal energy:
$$W_2 = \frac{n R (T_i - T_f)}{\gamma - 1}$$

Substitute our known values:
$$W_2 = \frac{1 \cdot R (T_0 - 2.25 T_0)}{5/3 - 1} = \frac{-1.25 R T_0}{2/3}$$
$$W_2 = -1.875 R T_0$$

The negative sign is the perfect conclusion. It tells us that the volume of the right chamber decreased, meaning work was done **ON** the gas, not by it. And with that, we've completely unraveled the mysteries of this thermodynamic system!

Visualized Solution

$Initial State$

$Final State$

$Adiabatic Process (Right Chamber)$

$Temperature T_{2}$

$Calculating T_{2}$

$Volume Constraint$

$Calculating V_{2}$

$Calculating V_{1}$

$Temperature T_{1}$

$Work Done (Right Chamber)$

$Final Calculation$

The Sigma Insight: Thermodynamic Processes

The Right Chamber

The Left Chamber

The Final Act

Similar Questions

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Two moles of an ideal monoatomic gas occupies a volume $V$ at $2 7^{\circ} C$ . The gas expands adiabatically to a volume $2 V$ . Calculate (i) the final temperature of the gas and (ii) change in its internal energy.

$n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Process (Right Chamber)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Volume Constraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating V2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating V1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done (Right Chamber)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation