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

Explanation

To solve the problem, we will use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when the pressure is held constant. The formula for Charles's Law can be expressed as:

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

Where:
- \( V_1 \) = Volume at temperature \( T_1 \)
- \( V_2 \) = Volume at temperature \( T_2 \)

### Step-by-Step Solution:

1. **Identify the given values:**
- Volume at ice-cold water \( V_1 = 47.5 \) units
- Temperature at ice-cold water \( T_1 = 0 \, ^\circ C = 273 \, K \)
- Volume at boiling liquid \( V_2 = 67 \) units
- We need to find \( T_2 \) (the boiling point of the liquid).

2. **Apply Charles's Law:**
Using the formula:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

3. **Rearranging the formula to solve for \( T_2 \):**
\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \]

4. **Substituting the known values into the equation:**
\[ T_2 = \frac{67 \cdot 273}{47.5} \]

5. **Calculating \( T_2 \):**
- First, calculate \( 67 \cdot 273 = 18291 \).
- Then divide by \( 47.5 \):
\[ T_2 = \frac{18291}{47.5} \approx 385 \, K \]

6. **Convert \( T_2 \) from Kelvin to Celsius:**
\[ T_2(°C) = T_2(K) - 273 = 385 - 273 = 112 \, °C \]

### Final Answer:
The boiling point of the liquid is \( 112 \, °C \).