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. (a)`135^(@)C`
B. (b)`125^(@)C`
C. (c)`112^(@)C`
D. (d)`100^(@)C`

Answer

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 pressure is held constant.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Volume when immersed in ice-cold water (V1) = 47.5 units
- Temperature of ice-cold water (T1) = 0°C = 273 K
- Volume when immersed in boiling liquid (V2) = 67 units
- Temperature of boiling liquid (T2) = ? (This is what we need to find)

2. **Apply Charles's Law:**
According to Charles's Law:
\[ \frac{V1}{T1} = \frac{V2}{T2} \]

3. **Substitute the Known Values:**
Plugging in the known values into the equation:
\[ \frac{47.5}{273} = \frac{67}{T2} \]

4. **Cross-Multiply to Solve for T2:**
Cross-multiplying gives us:
\[ 47.5 \cdot T2 = 67 \cdot 273 \]

5. **Calculate the Right Side:**
Calculate \(67 \cdot 273\):
\[ 67 \cdot 273 = 18291 \]

6. **Isolate T2:**
Now, isolate \(T2\):
\[ T2 = \frac{18291}{47.5} \]

7. **Perform the Division:**
Calculate:
\[ T2 \approx 385 K \]

8. **Convert Kelvin to Celsius:**
To convert Kelvin back to Celsius:
\[ T2 (°C) = T2 (K) - 273 = 385 - 273 = 112 °C \]

9. **Final Answer:**
The boiling point of the liquid is **112°C**.