What is the increase in volume, when the temperature of `600 mL` of air increases from `27^(@)C` to `47^(@)C` under constant pressure?

Options

A. `80 mL`
B. `40 mL`
C. `640 mL`
D. `500 mL`

Explanation

To solve the problem of finding the increase in volume when the temperature of 600 mL of air increases from 27°C to 47°C under constant pressure, we can use Charles's Law. Charles's Law 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. **Convert Temperatures from Celsius to Kelvin**:
- The initial temperature (T1) is 27°C.
- To convert to Kelvin:
\[ T1 = 27 + 273 = 300 \, K \]
- The final temperature (T2) is 47°C.
- To convert to Kelvin:
\[ T2 = 47 + 273 = 320 \, K \]

2. **Use Charles's Law**:
- According to Charles's Law:
\[ \frac{V1}{T1} = \frac{V2}{T2} \]
- Where:
- \( V1 = 600 \, mL \) (initial volume)
- \( T1 = 300 \, K \) (initial temperature)
- \( V2 \) = ? (final volume)
- \( T2 = 320 \, K \) (final temperature)

3. **Substitute Known Values into the Equation**:
- Plugging in the values we have:
\[ \frac{600}{300} = \frac{V2}{320} \]

4. **Cross Multiply to Solve for V2**:
- Cross multiplying gives:
\[ 600 \times 320 = 300 \times V2 \]
- This simplifies to:
\[ 192000 = 300 \times V2 \]

5. **Solve for V2**:
- Dividing both sides by 300:
\[ V2 = \frac{192000}{300} = 640 \, mL \]

6. **Calculate the Increase in Volume**:
- The increase in volume is given by:
\[ \text{Increase in Volume} = V2 - V1 = 640 \, mL - 600 \, mL = 40 \, mL \]

### Final Answer:
The increase in volume is **40 mL**.

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