An ideal gas expands at a constant pressure of 7.0 atm from 280 mL to 630 mL. Heat then flows out of the gas, at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate the total work done by the gas in the process

Answer

To solve the problem of calculating the total work done by the gas during its expansion and subsequent cooling, we can break it down into steps.

### Step-by-Step Solution:

1. **Identify the Process**: The gas undergoes two processes:
- **Expansion at Constant Pressure**: The gas expands from 280 mL to 630 mL at a constant pressure of 7.0 atm.
- **Cooling at Constant Volume**: The gas cools down at constant volume until the temperature reaches its original value.

2. **Calculate the Work Done During Expansion**:
The work done (W) by the gas during expansion at constant pressure is given by the formula:
\[ W = P \times \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

- Convert the volumes from mL to L:
- Initial Volume, \( V_i = 280 \, \text{mL} = 0.280 \, \text{L} \)
- Final Volume, \( V_f = 630 \, \text{mL} = 0.630 \, \text{L} \)

- Calculate the change in volume:
\[ \Delta V = V_f - V_i = 0.630 \, \text{L} - 0.280 \, \text{L} = 0.350 \, \text{L} \]

- Now, substitute the values into the work done formula:
\[ W = 7.0 \, \text{atm} \times 0.350 \, \text{L} = 2.45 \, \text{L} \cdot \text{atm} \]

3. **Consider the Cooling Process**:
During the cooling process at constant volume, the work done is zero because there is no change in volume (\( \Delta V = 0 \)). Therefore:
\[ W_{\text{cooling}} = 0 \]

4. **Total Work Done**:
The total work done by the gas in the entire process is simply the work done during the expansion, as no work is done during the cooling phase:
\[ W_{\text{total}} = W_{\text{expansion}} + W_{\text{cooling}} = 2.45 \, \text{L} \cdot \text{atm} + 0 = 2.45 \, \text{L} \cdot \text{atm} \]

### Final Answer:
The total work done by the gas in the process is **2.45 L·atm**.