Calculate the work done by 16g of oxygen gas (assume ideal behaviour) of molar mass 32 g `mol^(-1)` undergoing isothermal reversible expansion at 300 K from an initial volume of 2.5 L to the final volume of 25 L in litre atm.
(`R = 8.2 xx 10^(-2) L "atm K^(-1) "mol"^(-1)`)

Options

A. `-56.64`
B. `28.32`
C. `113.28`
D. `56.64`

Explanation

To calculate the work done by 16 g of oxygen gas undergoing isothermal reversible expansion, we can use the formula for work done in an isothermal process for an ideal gas:

\[ W = -2.303 \, nRT \, \log \left( \frac{V_2}{V_1} \right) \]

### Step 1: Calculate the number of moles (n)
Given:
- Mass of oxygen gas = 16 g
- Molar mass of oxygen gas = 32 g/mol

Using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Substitute the values into the work done formula
Given:
- R = \(8.2 \times 10^{-2} \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- T = 300 K
- \(V_1 = 2.5 \, \text{L}\)
- \(V_2 = 25 \, \text{L}\)

Now substituting the values into the work done formula:
\[ W = -2.303 \times n \times R \times T \times \log \left( \frac{V_2}{V_1} \right) \]
\[ W = -2.303 \times 0.5 \times (8.2 \times 10^{-2}) \times 300 \times \log \left( \frac{25}{2.5} \right) \]

### Step 3: Calculate the logarithmic term
\[ \frac{V_2}{V_1} = \frac{25}{2.5} = 10 \]
\[ \log(10) = 1 \]

### Step 4: Substitute the logarithmic value back into the equation
\[ W = -2.303 \times 0.5 \times (8.2 \times 10^{-2}) \times 300 \times 1 \]

### Step 5: Calculate the work done
Now, calculate the numerical values:
\[ W = -2.303 \times 0.5 \times 8.2 \times 10^{-2} \times 300 \]
\[ W = -2.303 \times 0.5 \times 24.6 \]
\[ W = -2.303 \times 12.3 = -28.32 \, \text{L atm} \]

### Final Answer
The work done by the gas is:
\[ W = -28.32 \, \text{L atm} \]