A constant pressure air thermometer gave a reading of `47.5` units of volume when immersed in ice cold water, and 67 units in a boiling liquid. The boiling point of the liquid will be

Options

A. `135^(@)C`
B. `125^(@)C`
C. `112^(@)C`
D. `100^(@)C`

Explanation

To solve the problem, we will use the relationship between the volume of gas and temperature at constant pressure, which can be expressed as:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where:
- \( V_1 \) = Volume at the ice point (47.5 units)
- \( T_1 \) = Temperature at the ice point (0°C = 273 K)
- \( V_2 \) = Volume at the boiling point (67 units)
- \( T_2 \) = Temperature at the boiling point (unknown, in °C)

### Step 1: Assign the known values
- \( V_1 = 47.5 \)
- \( T_1 = 0 + 273 = 273 \, \text{K} \)
- \( V_2 = 67 \)

### Step 2: Set up the equation
Using the formula, we can write:

\[ \frac{47.5}{273} = \frac{67}{T_2 + 273} \]

### Step 3: Cross-multiply to solve for \( T_2 \)
Cross-multiplying gives:

\[ 47.5 \cdot (T_2 + 273) = 67 \cdot 273 \]

### Step 4: Expand and simplify the equation
Expanding the left side:

\[ 47.5 T_2 + 47.5 \cdot 273 = 67 \cdot 273 \]

Calculating \( 67 \cdot 273 \):

\[ 67 \cdot 273 = 18291 \]

Calculating \( 47.5 \cdot 273 \):

\[ 47.5 \cdot 273 = 12997.5 \]

Now substituting these values back into the equation:

\[ 47.5 T_2 + 12997.5 = 18291 \]

### Step 5: Isolate \( T_2 \)
Subtract \( 12997.5 \) from both sides:

\[ 47.5 T_2 = 18291 - 12997.5 \]

Calculating the right side:

\[ 18291 - 12997.5 = 5293.5 \]

So we have:

\[ 47.5 T_2 = 5293.5 \]

### Step 6: Solve for \( T_2 \)
Now divide both sides by \( 47.5 \):

\[ T_2 = \frac{5293.5}{47.5} \approx 111.5 \, \text{°C} \]

### Final Answer
The boiling point of the liquid is approximately **111.5 °C**.

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