A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume of gas in two parts respectively are (P, 5V) and (10P, V). If now the piston is left free and the system undergoes isothermal process, then the volumes of the gas in two parts respectively are

Options

A. 4V, 2V
B. 5V , V
C. 2V , 4V
D. 3V , 3V

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
The gas cylinder is divided into two parts by a piston. The initial conditions are:
- Part 1: Pressure = P, Volume = 5V
- Part 2: Pressure = 10P, Volume = V

### Step 2: Total Volume Calculation
The total volume of the gas in the cylinder is the sum of the volumes of both parts:
\[ V_{\text{total}} = 5V + V = 6V \]

### Step 3: Isothermal Process Understanding
In an isothermal process, the temperature remains constant. Therefore, the product of pressure and volume for each part of the gas will remain constant:
\[ P_1 V_1 = P_2 V_2 \]
where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume after the piston is allowed to move.

### Step 4: Set Up the Equations
Let \(P'\) be the final pressure in both parts after equilibrium is reached. The new volumes will be:
- Volume of Part 1: \(V_1' = 6V - V_2'\)
- Volume of Part 2: \(V_2' = V_2\)

Using the isothermal condition for both parts:
1. For Part 1:
\[ PV_1 = P'V_1' \implies P(5V) = P'(6V - V_2') \]
2. For Part 2:
\[ 10PV = P'V_2' \]

### Step 5: Solve for Final Pressure
From the second equation:
\[ P' = \frac{10PV}{V_2'} \]

Substituting \(P'\) into the first equation:
\[ P(5V) = \frac{10PV}{V_2'}(6V - V_2') \]

### Step 6: Rearranging the Equation
Cancel \(P\) from both sides (assuming \(P \neq 0\)):
\[ 5V = \frac{10V(6V - V_2')}{V_2'} \]

Cross-multiplying gives:
\[ 5V \cdot V_2' = 10V(6V - V_2') \]

### Step 7: Simplifying the Equation
\[ 5V \cdot V_2' = 60V - 10V \cdot V_2' \]
\[ 5V \cdot V_2' + 10V \cdot V_2' = 60V \]
\[ 15V \cdot V_2' = 60V \]
\[ V_2' = 4V \]

### Step 8: Find Volume of Part 1
Now substitute \(V_2' = 4V\) back to find \(V_1'\):
\[ V_1' = 6V - V_2' = 6V - 4V = 2V \]

### Final Answer
The final volumes of the gas in the two parts are:
- Volume of Part 1: \(2V\)
- Volume of Part 2: \(4V\)

Thus, the answer is \(2V\) and \(4V\).