A glass bulb is connected to an open-limb manometer. The level of mercury on both limbs was same. The bulb was heated to `57^@C`. If the room temperature and atmospheric pressure are `27^@C` and 750 mmHg, the difference of levels in the two limbs now will be

Options

A. 2.5 cm
B. 5cm
C. 7.5 cm
D. 10 cm

Answer

To solve the problem, we will use the ideal gas law, specifically the relationship between pressure and temperature for a gas at constant volume. The relevant formula is:

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Where:
- \( P_1 \) = initial pressure
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( T_2 \) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = 57^\circ C = 57 + 273 = 330 \, K \)

### Step 2: Identify the initial pressure
- The initial pressure \( P_1 = 750 \, mmHg \)

### Step 3: Use the formula to find the final pressure \( P_2 \)
Substituting the known values into the formula:

\[ \frac{750 \, mmHg}{300 \, K} = \frac{P_2}{330 \, K} \]

### Step 4: Solve for \( P_2 \)
Cross-multiplying gives:

\[ P_2 = \frac{750 \, mmHg \times 330 \, K}{300 \, K} \]

Calculating \( P_2 \):

\[ P_2 = \frac{750 \times 330}{300} = 825 \, mmHg \]

### Step 5: Calculate the difference in mercury levels
The difference in levels of mercury in the two limbs of the manometer is:

\[ \text{Difference} = P_2 - P_1 = 825 \, mmHg - 750 \, mmHg = 75 \, mmHg \]

### Final Answer
The difference of levels in the two limbs now will be **75 mmHg**.

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