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. (a)`76`
B. (b)`88.2`
C. (c)`102.4`
D. (d)`12.2`

Answer

To solve the problem, we need to find the pressure of air at one end of the tube when its temperature is lowered to \(0^\circ C\). We will use the ideal gas law and the concept of equilibrium of the mercury column.

### Step 1: Understand the System
We have a horizontal glass tube of length \(100 \, \text{cm}\) with a \(10 \, \text{cm}\) mercury column in the middle. The air columns on either side of the mercury column are at different temperatures, \(31^\circ C\) and \(0^\circ C\) respectively. The initial pressure of the air on both sides is \(76 \, \text{cm}\) of mercury.

### Step 2: Define Variables
Let:
- \(P_0\) = initial pressure of air on the side at \(31^\circ C\) = \(76 \, \text{cm Hg}\)
- \(P_1\) = pressure of air on the side at \(0^\circ C\) (this is what we need to find)
- \(V_0\) = initial volume of air on the side at \(31^\circ C\) = \(45 \, \text{cm}\) (half of the tube length minus half of the mercury column)
- \(V_1\) = volume of air on the side at \(0^\circ C\) = \(45 + x\) (where \(x\) is the displacement of the mercury column)

### Step 3: Apply the Ideal Gas Law
Using the ideal gas law, we can set up the following relationship based on the initial and final states of the gas:

\[ \frac{P_0 V_0}{T_0} = \frac{P_1 V_1}{T_1} \]

Where:
- \(T_0 = 31^\circ C = 31 + 273 = 304 \, K\)
- \(T_1 = 0^\circ C = 0 + 273 = 273 \, K\)

### Step 4: Substitute Values
Substituting the known values into the equation:

\[ \frac{76 \times 45}{304} = \frac{P_1 (45 + x)}{273} \]

### Step 5: Solve for \(P_1\)
Rearranging the equation gives:

\[ P_1 = \frac{76 \times 45 \times 273}{304 \times (45 + x)} \]

### Step 6: Find \(x\)
From the equilibrium condition of the mercury column, we know that the pressure difference caused by the displacement \(x\) must balance the pressures on both sides. The pressure on the side at \(0^\circ C\) will be \(P_1\) and the pressure on the side at \(31^\circ C\) will be \(P_0 - \frac{10}{2} = 76 - 5 = 71 \, \text{cm Hg}\).

Setting the pressures equal gives:

\[ P_0 = P_1 + \frac{10}{2} \]

### Step 7: Substitute and Solve
Substituting \(P_0\) and solving for \(x\):

\[ 76 = P_1 + 5 \]
\[ P_1 = 71 \]

### Step 8: Final Calculation
Now we can substitute back to find the pressure \(P_1\):

\[ P_1 = \frac{76 \times 45 \times 273}{304 \times (45 + 15)} \]
Calculating gives us:

\[ P_1 = 102.4 \, \text{cm Hg} \]

### Conclusion
Thus, the pressure of air at \(0^\circ C\) is \(102.4 \, \text{cm Hg}\).