A thermally insulated piston divides a nonconducting container in two compartements, right compartement of `2V, T` and `2P`, while in the left compartement the respective values are `V, T` and `P`. Total moles in total system of both comartments is `5` moles. if piston can slide freely, and in the final equilibrium position, volume of right compartement is `(xV)/(5)` then find the value of `x`.

Answer

To solve the problem step by step, let's break down the information given and apply the ideal gas law for both compartments.

### Step 1: Understand the Initial Conditions
We have two compartments divided by a thermally insulated piston:
- Right compartment: Volume = \(2V\), Temperature = \(T\), Pressure = \(2P\)
- Left compartment: Volume = \(V\), Temperature = \(T\), Pressure = \(P\)
- Total moles in the system = \(5\) moles

### Step 2: Define Moles in Each Compartment
Let:
- \(n_1\) = number of moles in the right compartment
- \(n_2\) = number of moles in the left compartment

From the problem, we know:
\[ n_1 + n_2 = 5 \quad \text{(1)} \]

### Step 3: Apply the Ideal Gas Law
Using the ideal gas equation \(PV = nRT\), we can express the number of moles in each compartment.

For the right compartment:
\[ 2P \cdot 2V = n_1RT \implies n_1 = \frac{4PV}{RT} \quad \text{(2)} \]

For the left compartment:
\[ P \cdot V = n_2RT \implies n_2 = \frac{PV}{RT} \quad \text{(3)} \]

### Step 4: Relate the Moles
Now, we can find the ratio of moles \(n_1\) to \(n_2\) by dividing equation (2) by equation (3):
\[ \frac{n_1}{n_2} = \frac{\frac{4PV}{RT}}{\frac{PV}{RT}} = 4 \quad \text{(4)} \]

### Step 5: Solve for Moles
From equation (4), we have:
\[ n_1 = 4n_2 \]
Substituting this into equation (1):
\[ 4n_2 + n_2 = 5 \implies 5n_2 = 5 \implies n_2 = 1 \quad \text{and} \quad n_1 = 4 \quad \text{(5)} \]

### Step 6: Establish Equilibrium Conditions
At equilibrium, the pressures in both compartments will be equal:
\[ \frac{n_1RT}{V_1} = \frac{n_2RT}{V_2} \]
Since the temperature \(T\) and the gas constant \(R\) are the same, they cancel out:
\[ \frac{n_1}{V_1} = \frac{n_2}{V_2} \quad \text{(6)} \]

### Step 7: Express Volumes
Let \(V_1\) be the volume of the right compartment and \(V_2\) be the volume of the left compartment. From the problem, we know:
\[ V_1 + V_2 = 3V \quad \text{(since } 2V + V = 3V\text{)} \]
Substituting \(V_2 = 3V - V_1\) into equation (6):
\[ \frac{4}{V_1} = \frac{1}{3V - V_1} \]

### Step 8: Solve for \(V_1\)
Cross-multiplying gives:
\[ 4(3V - V_1) = V_1 \implies 12V - 4V_1 = V_1 \implies 12V = 5V_1 \implies V_1 = \frac{12V}{5} \quad \text{(7)} \]

### Step 9: Relate \(V_1\) to \(x\)
From the problem, we know that the volume of the right compartment is given as:
\[ V_1 = \frac{xV}{5} \]
Setting this equal to equation (7):
\[ \frac{xV}{5} = \frac{12V}{5} \]
Cancelling \(V\) from both sides:
\[ x = 12 \]

### Final Answer
Thus, the value of \(x\) is \(12\).

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