A vertical cylinder closed at both ends is fitted with a smooth piston dividing the volume into two parts each containing one mole of air. At the equilibrium temperature of 320 K, the volume of upper and lower parts are in the ratio 4:1. The ratio of volume will become 3:1 at a temperature of

Options

A. 450 K
B. 228 K
C. 420 K
D. 570 K

Explanation

To solve the problem step by step, we will use the principles of thermodynamics and the ideal gas law.

### Step 1: Understand the initial conditions
We have a vertical cylinder divided into two compartments by a piston. Each compartment contains one mole of air. The initial temperature is \( T_0 = 320 \, K \) and the volume ratio of the upper compartment to the lower compartment is \( V_1 : V_2 = 4 : 1 \).

### Step 2: Express the volumes in terms of a variable
Let the volume of the lower compartment be \( V_2 = V \). Then, the volume of the upper compartment can be expressed as:
\[ V_1 = 4V_2 = 4V \]

### Step 3: Set up the equation for the second scenario
In the second scenario, we want to find the temperature \( T' \) at which the volume ratio becomes \( V_1' : V_2' = 3 : 1 \). Let:
\[ V_2' = V \quad \text{(lower compartment)} \]
Then, the volume of the upper compartment becomes:
\[ V_1' = 3V_2' = 3V \]

### Step 4: Apply the ideal gas law
Using the ideal gas law, we know:
\[ PV = nRT \]
For the upper compartment at the initial state:
\[ P_1 V_1 = nRT_0 \quad \Rightarrow \quad P_1 (4V) = 1 \cdot R \cdot 320 \]
For the lower compartment at the initial state:
\[ P_2 V_2 = nRT_0 \quad \Rightarrow \quad P_2 (V) = 1 \cdot R \cdot 320 \]

### Step 5: Set up the pressure difference equations
From the equilibrium condition, we have:
\[ P_2 - P_1 = \frac{mg}{A} \quad \text{(1)} \]
For the second scenario:
\[ P_2' - P_1' = \frac{mg}{A} \quad \text{(2)} \]
From equations (1) and (2), we can equate the pressure differences.

### Step 6: Relate pressures and temperatures
Using the ideal gas law again for the new volumes:
\[ P_1' V_1' = nRT' \quad \Rightarrow \quad P_1' (3V) = 1 \cdot R \cdot T' \]
\[ P_2' V_2' = nRT' \quad \Rightarrow \quad P_2' (V) = 1 \cdot R \cdot T' \]

### Step 7: Solve for the new temperature
Using the relationships derived from the ideal gas law, we can express the pressures in terms of volumes and temperatures:
\[ \frac{P_1}{P_2} = \frac{T_0}{4V} \quad \text{and} \quad \frac{P_1'}{P_2'} = \frac{T'}{3V} \]
Equating the pressure differences and substituting the values:
\[ \frac{T_0}{4V} - \frac{T_0}{V} = \frac{T'}{3V} - \frac{T'}{V} \]

### Step 8: Simplify and solve for \( T' \)
After simplification, we find:
\[ T' = T_0 \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{3}{2} \]
Substituting \( T_0 = 320 \, K \):
\[ T' = 320 \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{3}{2} = 450 \, K \]

### Conclusion
Thus, the temperature at which the volume ratio becomes \( 3:1 \) is \( T' = 450 \, K \).

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