16 g of `O_(2)` gas at STP is expanded so that volume is doubled. Hence, work done is

Options

A. `-22.4` L atm
B. `-44.8` L atm
C. `-11.2` L atm
D. `-5.6` L atm

Answer

To solve the problem of calculating the work done when 16 g of O₂ gas is expanded at STP (Standard Temperature and Pressure), we will follow these steps:

### Step-by-Step Solution:

1. **Calculate the number of moles of O₂ gas:**
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molecular mass}} = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{moles} \]

2. **Determine the initial volume (V₁) of the gas at STP:**
At STP, 1 mole of gas occupies 22.4 liters. Therefore, for 0.5 moles:
\[ V_1 = 0.5 \, \text{moles} \times 22.4 \, \text{L/mole} = 11.2 \, \text{L} \]

3. **Calculate the final volume (V₂) after expansion:**
The volume is doubled, so:
\[ V_2 = 2 \times V_1 = 2 \times 11.2 \, \text{L} = 22.4 \, \text{L} \]

4. **Calculate the work done (W) during the expansion:**
The work done during expansion at constant pressure is given by:
\[ W = P \times \Delta V \]
where \(\Delta V = V_2 - V_1\). Since the pressure at STP is 1 atm:
\[ \Delta V = V_2 - V_1 = 22.4 \, \text{L} - 11.2 \, \text{L} = 11.2 \, \text{L} \]
Therefore, the work done is:
\[ W = 1 \, \text{atm} \times 11.2 \, \text{L} = 11.2 \, \text{L atm} \]

5. **Determine the sign of the work done:**
Since this is an expansion, the work done by the system is considered negative:
\[ W = -11.2 \, \text{L atm} \]

### Final Answer:
The work done during the expansion is \(-11.2 \, \text{L atm}\).