A container is divided into two chambers by a partition. The volume of first chamber is 4.5 L and se

Q: Why did we assume the temperature is constant?

The problem states the mixture 'attains equilibrium' without mentioning any heat exchange or initial temperature differences. In such standard problems, it implies thermal equilibrium at the same initial temperature.

Q: Can we just average the pressures?

No! Pressure is an intensive property. You must use the weighted average based on volume, which is exactly what the $p_1V_1 + p_2V_2 = p(V_1 + V_2)$ equation does.

Q: What happened to the number of moles given in the question?

The number of moles ($3.0$ mol and $4.0$ mol) are actually extra information! Since we know $p$ and $V$ for both chambers and assume constant $T$, the ratio of moles is already fixed by $pV$. You don't strictly need them to solve for the final pressure.

Q: What if the gases react with each other?

If the gases react, the total number of moles might change, and the reaction might release or absorb heat, changing the temperature. Here, we assume they are non-reacting ideal gases.

Q: Why write it as $25.5 \times 10^{-1}$ instead of $2.55$?

This is a common trick in numerical answer type questions in JEE to ensure students enter an integer or a specific decimal format. Always read the requested format carefully!

Question

A container is divided into two chambers by a partition. The volume of first chamber is $4.5$ L and second chamber is $5.5$ L. The first chamber contain $3.0$ mol of gas at pressure $2.0$ atm and second chamber contain $4.0$ mol of gas at pressure $3.0$ atm. After the partition is removed and the mixture attains equilibrium, then the common equilibrium pressure existing in the mixture is $x 	imes 10^{-1}$ atm. Value of $x$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a rigid container divided into two separate chambers by a partition. The first chamber has a volume of $4.5	ext{ L}$ and holds gas at a pressure of $2.0	ext{ atm}$. The second chamber is slightly larger at $5.5	ext{ L}$, with a pressure of $3.0	ext{ atm}$. We are about to remove this partition and let the gases mix.

When we remove the partition, the gases will mix and reach a new equilibrium. The total number of moles of gas remains conserved. Since the problem doesn't mention any temperature change, we assume the temperature remains constant. According to the ideal gas law, $pV = nRT$, the number of moles is directly proportional to the product of pressure and volume.

### The Master Equation
This gives us a beautiful and simple relation: the sum of the initial $pV$ products equals the final $pV$ product. Let's substitute our known values into this equation. We have $2.0$ times $4.5$ for the first chamber, plus $3.0$ times $5.5$ for the second chamber. This total must equal the final unknown pressure $p$ times the total combined volume, which is $4.5$ plus $5.5$.

$$p_1V_1 + p_2V_2 = p(V_1 + V_2)$$

$$(2.0)(4.5) + (3.0)(5.5) = p(4.5 + 5.5)$$

### Final Calculation
Let's do the math. $2.0$ multiplied by $4.5$ is exactly $9.0$. And $3.0$ multiplied by $5.5$ gives us $16.5$. On the right side, the total volume is a nice, round $10.0	ext{ L}$. Adding $9.0$ and $16.5$, we get a total of $25.5$ on the left side. So, $25.5$ equals $10$ times $p$.

$$9.0 + 16.5 = p(10.0)$$

$$25.5 = 10p$$

To find the final equilibrium pressure $p$, we simply divide $25.5$ by $10$. This gives us a final pressure of $2.55	ext{ atm}$. Notice how the final pressure is somewhere between the two initial pressures, which makes perfect physical sense.

$$p = 2.55 	ext{ atm}$$

We are almost done, but we need to format our answer exactly as the question asks. The question wants the pressure in the form of $x 	imes 10^{-1}$. We can rewrite $2.55$ as $25.5 	imes 10^{-1}$. Comparing this with the given format, we can clearly see that the value of $x$ is **$25.5$**. That's our final answer!

Enter Numerical Value:

Visualized Solution

Initial State of the Container

Conservation of Moles

Equation for Final Pressure

Calculating $p V$ Products

Solving for $p$

Formatting the Answer

What if Temperatures Differed?

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup Imagine a rigid container divided into two separate chambers by a partition

The Master Equation This gives us a beautiful and simple relation: the sum of the initial $p V$ products equals the final $p V$ product

Final Calculation Let's do the math. $2.0$ multiplied by $4.5$ is exactly $9.0$

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