The work done when `1.6 xx 10^(-2) ` kg of oxygen at 300 K are expended isothermally and reversibly , till its volume is tripled is `(R=8.314 JK^(-1) mol^(-1))`

Options

A. `- 0.646 KJ`
B. `-1.370 kJ`
C. `-864.5 kJ`
D. `-645.4 kJ`

Explanation

To solve the problem of calculating the work done when 1.6 x 10^(-2) kg of oxygen is expanded isothermally and reversibly until its volume is tripled, we can follow these steps:

### Step 1: Convert mass of oxygen to moles
We need to find the number of moles of oxygen (O₂). The molar mass of oxygen is approximately 32 g/mol.

Given:
- Mass of oxygen = 1.6 x 10^(-2) kg = 16 g

Using the formula:
\[ \text{Number of moles (n)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ n = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 2: Identify initial and final volumes
Let the initial volume \( V_1 = V \) and the final volume \( V_2 = 3V \) since the volume is tripled.

### Step 3: Use the work done formula for isothermal expansion
The work done \( W \) during an isothermal and reversible expansion can be calculated using the formula:
\[ W = -2.303 \, nRT \log\left(\frac{V_2}{V_1}\right) \]

### Step 4: Substitute the known values into the formula
Given:
- \( n = 0.5 \) moles
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( T = 300 \, \text{K} \)
- \( V_2 = 3V \) and \( V_1 = V \)

Now substituting these values into the work done formula:
\[ W = -2.303 \times 0.5 \times 8.314 \times 300 \log\left(\frac{3V}{V}\right) \]
\[ W = -2.303 \times 0.5 \times 8.314 \times 300 \log(3) \]

### Step 5: Calculate \( \log(3) \)
Using a calculator, we find:
\[ \log(3) \approx 0.477 \]

### Step 6: Calculate the work done
Now substituting \( \log(3) \) back into the equation:
\[ W = -2.303 \times 0.5 \times 8.314 \times 300 \times 0.477 \]

Calculating step-by-step:
1. \( 0.5 \times 8.314 = 4.157 \)
2. \( 4.157 \times 300 = 1247.1 \)
3. \( 1247.1 \times 0.477 \approx 595.5 \)
4. \( W = -2.303 \times 595.5 \approx -1370.1 \, \text{J} \)

### Step 7: Convert to kilojoules
Since the answer is typically given in kilojoules:
\[ W \approx -1.370 \, \text{kJ} \]

### Final Answer
The work done is approximately \( -1.370 \, \text{kJ} \).

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