A horizontal uniform glass tube of `100cm` length is sealed at both ends contains `10 cm` mercury column in the middle , the temperature and pressure of air on either side of mercury column are repectively `31^(@)C` and `76cm` of mercury, if the air column at one end is kept at` 0^(@)C` and the other end at `273^(@)C` then pressure if air which is `0^(@)C` is
(in `cm` of `Hg`)

Options

A. `76`
B. `88.2`
C. `102.4`
D. `122`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
We have a horizontal glass tube of length 100 cm, sealed at both ends, containing a 10 cm mercury column in the middle. The air pressure on either side of the mercury column is 76 cm of mercury at a temperature of 31°C.

### Step 2: Define the Lengths
Since the total length of the tube is 100 cm and there is a 10 cm mercury column in the middle, the lengths of the air columns on either side of the mercury column are:
- Length of air column on the left (L1) = 45 cm
- Length of air column on the right (L2) = 45 cm

### Step 3: Identify the New Conditions
One end of the tube is kept at 0°C and the other at 273°C. This means:
- Temperature on the left side (T1) = 0°C = 273 K
- Temperature on the right side (T2) = 273°C = 546 K

### Step 4: Apply the Gas Law
Using the ideal gas law for both sides of the mercury column, we can set up the following relationship based on the fact that the number of moles of air remains constant:

\[ \frac{P_1 \cdot L_1}{T_1} = \frac{P_2 \cdot L_2}{T_2} \]

Where:
- \(P_1\) = pressure of air at 0°C (to be found)
- \(P_2\) = pressure of air at 273°C = 76 cm of mercury
- \(L_1\) = length of air column at 0°C = \(45 - x\)
- \(L_2\) = length of air column at 273°C = \(45 + x\)

### Step 5: Determine the Shift of Mercury Column
Since the mercury column shifts due to the temperature difference, let \(x\) be the distance the mercury column moves towards the cooler side (0°C). Thus:
- New length on the left (L1) = \(45 - x\)
- New length on the right (L2) = \(45 + x\)

### Step 6: Set Up the Equation
From the gas law, we have:
\[ \frac{P_1 (45 - x)}{273} = \frac{76 (45 + x)}{546} \]

### Step 7: Solve for x
Cross-multiplying gives:
\[ P_1 (45 - x) \cdot 546 = 76 (45 + x) \cdot 273 \]

### Step 8: Simplify the Equation
Since the pressures are equal, we can simplify and solve for \(x\):
\[ P_1 (45 - x) = \frac{76 \cdot 273}{546} (45 + x) \]

### Step 9: Calculate x
After solving the above equation, we find \(x = 15\) cm.

### Step 10: Calculate the Pressure \(P_1\)
Now substituting \(x\) back into the equation to find \(P_1\):
\[ P_1 \cdot (45 - 15) = \frac{76 \cdot 273}{546} \cdot (45 + 15) \]
\[ P_1 \cdot 30 = \frac{76 \cdot 273}{546} \cdot 60 \]
\[ P_1 = \frac{76 \cdot 273 \cdot 30}{546 \cdot 60} \]
Calculating this gives:
\[ P_1 = 102.3 \text{ cm of Hg} \]

### Final Answer
The pressure of air at 0°C is approximately **102.3 cm of Hg**.

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