16g of `O_(2)` at `28^(@)C` is compressed to half of its initial volume under reversible isothermal conditions. If this gas behaves ideally, the workdone and change in internal energy in this process are

Options

A. `Delta U = 0` and w = -1251.2 ln 2 J
B. `Delta U = 0` and w = +1251.2 ln 2J
C. `w = Delta U = +1251.2 ln 2J`
D. `Delta U = -1251.2 ln 2J` and `w = +1251.2 ln 2J`

Explanation

To solve the problem, we need to determine the work done and the change in internal energy when 16 g of \( O_2 \) is compressed to half its initial volume under reversible isothermal conditions.

### Step 1: Determine the number of moles of \( O_2 \)

The molar mass of \( O_2 \) is approximately 32 g/mol.

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

### Step 2: Identify the temperature in Kelvin

The temperature is given as \( 28^\circ C \). To convert this to Kelvin:

\[ T = 28 + 273 = 301 \, \text{K} \]

### Step 3: Determine the change in internal energy

For an isothermal process involving an ideal gas, the change in internal energy (\( \Delta U \)) is zero because internal energy is a function of temperature, and the temperature remains constant.

\[ \Delta U = 0 \]

### Step 4: Calculate the work done during the compression

The work done (\( W \)) on an ideal gas during an isothermal process can be calculated using the formula:

\[ W = -nRT \ln \left(\frac{V_2}{V_1}\right) \]

Since the gas is compressed to half its initial volume, we have \( V_2 = \frac{V_1}{2} \). Thus, the equation becomes:

\[ W = -nRT \ln \left(\frac{1/2}{1}\right) = -nRT \ln \left(\frac{1}{2}\right) = nRT \ln(2) \]

### Step 5: Substitute the values into the work done formula

Now, substituting the values of \( n \), \( R \), and \( T \):

- \( n = 0.5 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 301 \, \text{K} \)

\[ W = 0.5 \times 8.314 \times 301 \times \ln(2) \]

Calculating \( \ln(2) \approx 0.693 \):

\[ W = 0.5 \times 8.314 \times 301 \times 0.693 \]

\[ W \approx 0.5 \times 8.314 \times 301 \times 0.693 \approx 1251.2 \, \text{J} \]

### Final Results

1. **Change in Internal Energy (\( \Delta U \))**: 0 J
2. **Work Done (W)**: Approximately 1251.2 J

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