An insulated container of gas has two chambers separated by an insulating partition. One of the cham

Q: Why do we assume the degrees of freedom $f$ are the same for both gases?

The problem simply states 'ideal gas' for both chambers without specifying different types (like monatomic vs. diatomic). In such general cases, we assume they are of the same atomicity, allowing $f$ to cancel out.

Q: Can we just take the average of the two temperatures?

No, temperature is an intensive property and cannot be simply averaged unless the heat capacities and moles are identical. You must average the extensive property, which is the total internal energy.

Q: What if the container wasn't insulated?

If it wasn't insulated, heat would be exchanged with the surroundings ($\Delta Q \neq 0$). The final temperature would then depend on the temperature of the surroundings and the rate of heat transfer.

Q: Why did we use moles instead of mass in the internal energy formula?

The internal energy formula $U = \frac{f}{2}nRT$ is most directly expressed in moles. Using mass would require knowing the specific molar masses of the gases, which are not provided in the problem.

Question

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $p_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $p_2$ and temperature $T_2$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Options:

(A) $\frac{T_1 T_2 (p_1 V_1 + p_2 V_2)}{p_1 V_1 T_2 + p_2 V_2 T_1}$

(B) $\frac{p_1 V_1 T_1 + p_2 V_2 T_2}{p_1 V_1 + p_2 V_2}$

(C) $\frac{p_1 V_1 T_2 + p_2 V_2 T_1}{p_1 V_1 + p_2 V_2}$

(D) $\frac{T_1 T_2 (p_1 V_1 + p_2 V_2)}{p_1 V_1 T_1 + p_2 V_2 T_2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a rigid, perfectly insulated container divided into two distinct chambers by a partition. On the left side, we have an ideal gas characterized by pressure $p_1$, volume $V_1$, and temperature $T_1$. On the right side, we have another ideal gas at pressure $p_2$, volume $V_2$, and temperature $T_2$.

The core event of this problem is the removal of the partition. When this happens, the gases mix and eventually settle at a new, common equilibrium temperature, $T$. Our goal is to find an expression for this final temperature.

### The Master Equation: First Law of Thermodynamics
To solve this, we must rely on the First Law of Thermodynamics, which states:

$$\Delta Q = \Delta U + \Delta W$$

Let's analyze the constraints given in the problem:
1. **Insulated Container:** The walls do not allow any heat to enter or escape the system. Therefore, the heat exchanged is zero: $\Delta Q = 0$.
2. **No Work Done:** The partition is removed without any external force acting over a distance, and the rigid container's total volume remains constant. Thus, the work done is zero: $\Delta W = 0$.

Plugging these into the First Law, we get:

$$0 = \Delta U + 0 \implies \Delta U = 0$$

This is a profound realization: **The total internal energy of the system is strictly conserved.** The initial internal energy before mixing must equal the final internal energy after mixing ($U_i = U_f$).

### Formulating the Internal Energies
The internal energy of an ideal gas is given by $U = \frac{f}{2}nRT$, where $f$ is the degrees of freedom and $n$ is the number of moles.

Let's write down the initial internal energy, which is simply the sum of the energies of the two separate chambers. Assuming both gases are of the same type (e.g., both monatomic or both diatomic), they share the same degrees of freedom, $f$.

$$U_i = U_1 + U_2 = \frac{f}{2}n_1 R T_1 + \frac{f}{2}n_2 R T_2$$

After the partition is removed, the gases mix. The total number of moles becomes $(n_1 + n_2)$, and they reach a common temperature $T$. The final internal energy is:

$$U_f = \frac{f}{2}(n_1 + n_2) R T$$

### Equating and Simplifying
By conservation of internal energy, we equate $U_i$ and $U_f$:

$$\frac{f}{2}(n_1 + n_2) R T = \frac{f}{2}n_1 R T_1 + \frac{f}{2}n_2 R T_2$$

Notice how beautifully the $\frac{f}{2}$ and $R$ terms cancel out from both sides. This leaves us with a clean relationship for the final temperature:

$$(n_1 + n_2) T = n_1 T_1 + n_2 T_2$$

$$T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$$

### Final Calculation: Bringing in Pressure and Volume
Our expression for $T$ is in terms of moles, but the options provided are in terms of pressure and volume. We need to bridge this gap using the Ideal Gas Law, $pV = nRT$, which rearranges to $n = \frac{pV}{RT}$.

Let's substitute $n_1$ and $n_2$ for both chambers:

$$n_1 = \frac{p_1 V_1}{R T_1} \quad 	ext{and} \quad n_2 = \frac{p_2 V_2}{R T_2}$$

Plugging these into our temperature equation:

$$T = \frac{\left(\frac{p_1 V_1}{R T_1}ight) T_1 + \left(\frac{p_2 V_2}{R T_2}ight) T_2}{\frac{p_1 V_1}{R T_1} + \frac{p_2 V_2}{R T_2}}$$

In the numerator, the temperatures $T_1$ and $T_2$ cancel out perfectly:

$$T = \frac{\frac{p_1 V_1}{R} + \frac{p_2 V_2}{R}}{\frac{p_1 V_1}{R T_1} + \frac{p_2 V_2}{R T_2}}$$

We can factor out and cancel $\frac{1}{R}$ from both the numerator and the denominator:

$$T = \frac{p_1 V_1 + p_2 V_2}{\frac{p_1 V_1 T_2 + p_2 V_2 T_1}{T_1 T_2}}$$

Finally, bringing the common denominator $T_1 T_2$ up to the numerator yields our final, elegant result:

**$$T = \frac{T_1 T_2 (p_1 V_1 + p_2 V_2)}{p_1 V_1 T_2 + p_2 V_2 T_1}$$**

This matches option (a) perfectly. Always remember, when dealing with insulated mixing problems, conservation of internal energy is your ultimate master key!

Select Answer:

Visualized Solution

Initial Setup

First Law of Thermodynamics

Applying Constraints

Initial Internal Energy

Final Internal Energy

Conservation of Internal Energy

Number of Moles

Substituting Moles

Simplifying the Expression

What if the gases were different?

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup Imagine a rigid, perfectly insulated container divided into two distinct chambers by a partition

The Master Equation

Formulating the Internal Energies

Equating and Simplifying

Final Calculation

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Select Answer:

Visualized Solution

Initial Setup

First Law of Thermodynamics

Applying Constraints

Initial Internal Energy

Final Internal Energy

Conservation of Internal Energy

Number of Moles

Substituting Moles

Simplifying the Expression

What if the gases were different?

The Sigma Insight: First Law of Thermodynamics

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The Master Equation

Formulating the Internal Energies

Equating and Simplifying

Final Calculation

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