A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The p

Q: Why do we assume the number of moles is constant?

The problem states it is a 'closed tank'. This means no gas can enter or leave the system. Therefore, the total amount of gas, and the amount in each compartment (since the partition doesn't leak), must remain constant.

Q: Why are the final pressures equal?

The new partition can 'slide'. It will keep moving as long as there is a pressure difference pushing it. It only stops moving (reaches mechanical equilibrium) when the pressure on both sides is exactly the same.

Q: Why are the final temperatures equal?

The new partition can 'conduct heat'. Heat will naturally flow from the hotter gas to the colder gas until there is no longer a temperature difference. This state is called thermal equilibrium.

Q: What if the partition was adiabatic (insulating) but still movable?

In that case, the final pressures would still equalize, but the final temperatures would not. You would need to use the First Law of Thermodynamics (equating the work done by one gas to the work done on the other) to find the final state.

Q: Why did we use the relation $V \propto n$?

From the Ideal Gas Law, $V = \frac{nRT}{P}$. At final equilibrium, both $T$ and $P$ are the same for both compartments. Since $R$, $T$, and $P$ are all constants, the volume $V$ depends only on the number of moles $n$.

Question

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $	ext{m}^3$) of the compartment A after the system attains equilibrium is_______.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Compartments

Imagine a sturdy, closed tank. Inside this tank, we have two distinct compartments, A and B, both filled with oxygen gas. Initially, these two compartments are separated by a rigid, thick wall—a partition that is completely fixed and acts as a perfect heat insulator.

This means the gas in compartment A is completely isolated from the gas in compartment B. They can't mix, they can't push against each other, and they can't even share heat.

Let's look at the initial conditions given to us. Compartment A has a volume of $1\,	ext{m}^3$, a pressure of $5\,	ext{bar}$, and a temperature of $400\,	ext{K}$. Compartment B is larger but cooler and at a lower pressure, with a volume of $3\,	ext{m}^3$, a pressure of $1\,	ext{bar}$, and a temperature of $300\,	ext{K}$.

### The Unchanging Reality: Conservation of Moles

Before we even think about changing the partition, we need to identify what remains constant in this system. The tank is completely closed. No gas is being pumped in, and no gas is leaking out.

This brings us to a fundamental principle: the **conservation of moles**. The number of moles of oxygen in compartment A ($n_A$) and the number of moles in compartment B ($n_B$) will remain strictly constant, no matter what happens inside the tank.

We can calculate these initial moles using the Ideal Gas Law, $PV = nRT$. Rearranging for moles, we get $n = \frac{PV}{RT}$.

For compartment A:
$$n_A = \frac{5 	imes 1}{R 	imes 400} = \frac{5}{400R}$$

For compartment B:
$$n_B = \frac{1 	imes 3}{R 	imes 300} = \frac{3}{300R}$$

We don't need to plug in the value of the universal gas constant $R$ just yet. Keeping it as a variable will make our calculations much cleaner later on.

### The Game Changer: A New Partition

Now, the problem introduces a fascinating twist. The old, stubborn partition is removed and replaced with a new one. This new partition has two special properties: it can **slide**, and it can **conduct heat**. However, it still doesn't allow any gas to leak across.

What do these properties physically mean for our system?

First, the ability to slide means the partition will move until the forces on both sides are balanced. In the language of thermodynamics, this is called **mechanical equilibrium**. It guarantees that the final pressure in compartment A ($P_A'$) will be exactly equal to the final pressure in compartment B ($P_B'$). Let's call this common final pressure $P_f$.

Second, the ability to conduct heat means thermal energy will flow from the hotter gas to the colder gas until their temperatures equalize. This is **thermal equilibrium**. It guarantees that the final temperature in compartment A ($T_A'$) will be exactly equal to the final temperature in compartment B ($T_B'$). Let's call this common final temperature $T_f$.

### The Master Equation: Volume and Moles

At this final equilibrium state, both compartments share the exact same pressure ($P_f$) and the exact same temperature ($T_f$).

Let's look at the Ideal Gas Law again, but this time, let's solve for volume:
$$V = \frac{nRT_f}{P_f}$$

Notice something beautiful here? For both compartments, $R$, $T_f$, and $P_f$ are identical constants. This means the volume of a compartment is directly proportional to the number of moles it contains!

$$V \propto n$$

We can write this as a powerful ratio:
$$\frac{V_A'}{n_A} = \frac{V_B'}{n_B}$$

### Setting Up the Math

We know the total volume of the tank is fixed at $1\,	ext{m}^3 + 3\,	ext{m}^3 = 4\,	ext{m}^3$.

Let's assume the partition slides such that the volume of compartment A increases by an amount $x$. 
So, the new volume of compartment A is:
$$V_A' = 1 + x$$

Consequently, the new volume of compartment B must decrease by the same amount $x$:
$$V_B' = 3 - x$$

Now, we substitute these new volumes and our previously calculated moles into our master ratio:
$$\frac{1 + x}{\frac{5}{400R}} = \frac{3 - x}{\frac{3}{300R}}$$

### The Final Calculation

This equation might look a bit intimidating with those fractions in the denominator, but it simplifies beautifully.

First, let's simplify the fractions:
$\frac{5}{400R}$ simplifies to $\frac{1}{80R}$.
$\frac{3}{300R}$ simplifies to $\frac{1}{100R}$.

Substituting these back in:
$$\frac{1 + x}{\frac{1}{80R}} = \frac{3 - x}{\frac{1}{100R}}$$

This is equivalent to:
$$80R(1 + x) = 100R(3 - x)$$

The $R$ cancels out perfectly from both sides. We can also divide both sides by 20 to make the numbers even smaller:
$$4(1 + x) = 5(3 - x)$$

Now, we just expand the brackets:
$$4 + 4x = 15 - 5x$$

Bring all the $x$ terms to one side and the constants to the other:
$$4x + 5x = 15 - 4$$
$$9x = 11$$
$$x = \frac{11}{9}\,	ext{m}^3$$

We found $x$, the change in volume! But the question asks for the final volume of compartment A.

$$V_A' = 1 + x = 1 + \frac{11}{9}$$
$$V_A' = \frac{9}{9} + \frac{11}{9} = \frac{20}{9}\,	ext{m}^3$$

Converting this fraction to a decimal gives us our final, elegant answer:
$$V_A' \approx 2.22\,	ext{m}^3$$

### A Thought Experiment

Before we finish, let's ponder a variation of this problem. What if the new partition could slide, but it was still a perfect heat insulator (adiabatic)?

In that scenario, the pressures would still equalize ($P_A' = P_B'$), but the temperatures would not necessarily be equal ($T_A' 
eq T_B'$). You couldn't use the simple $V \propto n$ ratio. Instead, you would have to dive into the First Law of Thermodynamics, analyzing the work done by the expanding gas on the compressing gas. This is a classic, higher-level concept often tested in advanced exams, so always pay close attention to the specific words used to describe the partition!

Enter Numerical Value:

Visualized Solution

$I ni t ia l S t a t e A na l y s i s$

$C o n ser v a t i o n o f M o l es$

$C a l c u l a t in g I ni t ia l M o l es$

$F ina l E q u i l ib r i u m S t a t e$

$V o l u m e - M o l e R e l a t i o n s hi p$

$S e tt in g u p t h e E q u a t i o n$

$S im pl i f y in g t h e E q u a t i o n$

$S o l v in g f or x$

$F ina l V o l u m e o f C o m p a r t m e n t A$

$C o n ce pt u a l E x t e n s i o n$

The Sigma Insight: Gaseous State

The Setup

The Unchanging Reality

The Game Changer

The Master Equation

Setting Up the Math

The Final Calculation

A Thought Experiment

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Equation

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

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

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

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