`400 cm^(3)` of oxygen at `27^(@)C` were cooled to `-3^(@)C` without change in pressure. The contraction in volume will be as per Charle's law?

Options

A. 40
B. 30
C. 44.4
D. 360

Answer

To solve the problem using Charles's Law, we will follow these steps:

### Step 1: Convert Temperatures to Kelvin
We need to convert the temperatures from Celsius to Kelvin because gas laws use absolute temperature.

- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = -3^\circ C = -3 + 273 = 270 \, K \)

### Step 2: Write Down Charles's Law
Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature. Mathematically, it can be expressed as:

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

Where:
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature
- \( V_2 \) = final volume
- \( T_2 \) = final temperature

### Step 3: Substitute Known Values
We know:
- \( V_1 = 400 \, cm^3 \)
- \( T_1 = 300 \, K \)
- \( T_2 = 270 \, K \)

Substituting these values into the equation:

\[ \frac{400}{300} = \frac{V_2}{270} \]

### Step 4: Solve for \( V_2 \)
Cross-multiplying to solve for \( V_2 \):

\[ 400 \cdot 270 = 300 \cdot V_2 \]

Calculating the left side:

\[ 108000 = 300 \cdot V_2 \]

Now, divide both sides by 300 to find \( V_2 \):

\[ V_2 = \frac{108000}{300} = 360 \, cm^3 \]

### Step 5: Calculate the Contraction in Volume
The contraction in volume is the difference between the initial and final volumes:

\[ \text{Contraction} = V_1 - V_2 = 400 \, cm^3 - 360 \, cm^3 = 40 \, cm^3 \]

### Final Answer
The contraction in volume is \( 40 \, cm^3 \).

---