Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas?
(neglect the sign)

Answer

To solve the problem of calculating the work done by the gas during its expansion, we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done (W) by a gas during an isothermal expansion against a constant external pressure can be calculated using the formula:
\[ W = -P_{\text{ext}} \Delta V \]
where \( P_{\text{ext}} \) is the external pressure and \( \Delta V \) is the change in volume.

### Step 2: Convert Pressures to the Same Units
We need to convert the pressures given in bar to atm, as the final answer is required in L·atm. The conversion factor is:
1 bar = 0.98692 atm.

- Initial pressure \( P_i = 8.21 \, \text{bar} = 8.21 \times 0.98692 \, \text{atm} \approx 8.11 \, \text{atm} \)
- Final pressure \( P_f = 2.73 \, \text{bar} = 2.73 \times 0.98692 \, \text{atm} \approx 2.69 \, \text{atm} \)
- External pressure \( P_{\text{ext}} = 1 \, \text{bar} = 1 \times 0.98692 \, \text{atm} \approx 0.98692 \, \text{atm} \)

### Step 3: Calculate the Initial and Final Volumes
Using the Ideal Gas Law:
\[ PV = nRT \]
where:
- \( n = 2 \, \text{moles} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 300 \, \text{K} \)

Calculate the initial volume \( V_i \):
\[ V_i = \frac{nRT}{P_i} = \frac{2 \times 0.0821 \times 300}{8.11} \]

Calculating \( V_i \):
\[ V_i \approx \frac{49.26}{8.11} \approx 6.08 \, \text{L} \]

Calculate the final volume \( V_f \) using the final pressure \( P_f \):
\[ V_f = \frac{nRT}{P_f} = \frac{2 \times 0.0821 \times 300}{2.69} \]

Calculating \( V_f \):
\[ V_f \approx \frac{49.26}{2.69} \approx 18.32 \, \text{L} \]

### Step 4: Calculate the Change in Volume
\[ \Delta V = V_f - V_i = 18.32 \, \text{L} - 6.08 \, \text{L} = 12.24 \, \text{L} \]

### Step 5: Calculate the Work Done
Now, substituting the values into the work done formula:
\[ W = -P_{\text{ext}} \Delta V \]
\[ W = -0.98692 \, \text{atm} \times 12.24 \, \text{L} \]

Calculating \( W \):
\[ W \approx -12.06 \, \text{L·atm} \]

### Step 6: Neglect the Sign
Since we are asked to neglect the sign, the work done by the gas is:
\[ W \approx 12.06 \, \text{L·atm} \]

### Final Answer
The work done by the gas is approximately **12.06 L·atm**.

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