To solve the problem step by step, we will calculate the work done by the gas during its expansion. Here’s how to approach it:
### Step 1: Identify the given values
- Number of moles of gas (n) = 2 moles
- Initial pressure (P1) = 8.21 bar
- Final pressure (P2) = 2.73 bar
- External pressure (P_ext) = 1 bar
- Temperature (T) = 300 K
### Step 2: Convert pressures from bar to atm
To convert bar to atm, we use the conversion factor:
1 bar = 0.987 atm
- Initial pressure in atm (P1):
\[
P1 = 8.21 \, \text{bar} \times 0.987 \, \text{atm/bar} = 8.11 \, \text{atm}
\]
- Final pressure in atm (P2):
\[
P2 = 2.73 \, \text{bar} \times 0.987 \, \text{atm/bar} = 2.69 \, \text{atm}
\]
### Step 3: Calculate the initial and final volumes using the Ideal Gas Law
Using the Ideal Gas Law, \( PV = nRT \), we can find the volumes \( V1 \) and \( V2 \).
- For initial volume \( V1 \):
\[
V1 = \frac{nRT}{P1}
\]
Where \( R = 0.0821 \, \text{L atm/(K mol)} \)
Substituting the values:
\[
V1 = \frac{2 \, \text{mol} \times 0.0821 \, \text{L atm/(K mol)} \times 300 \, \text{K}}{8.11 \, \text{atm}} \approx 6.08 \, \text{L}
\]
- For final volume \( V2 \):
\[
V2 = \frac{nRT}{P2}
\]
Substituting the values:
\[
V2 = \frac{2 \, \text{mol} \times 0.0821 \, \text{L atm/(K mol)} \times 300 \, \text{K}}{2.69 \, \text{atm}} \approx 18.31 \, \text{L}
\]
### Step 4: Calculate the change in volume \( \Delta V \)
\[
\Delta V = V2 - V1 = 18.31 \, \text{L} - 6.08 \, \text{L} = 12.23 \, \text{L}
\]
### Step 5: Calculate the work done by the gas
The work done by the gas during expansion against a constant external pressure is given by:
\[
W = -P_{\text{ext}} \Delta V
\]
Since we are neglecting the sign, we will calculate the absolute value.
Substituting the values:
\[
W = 1 \, \text{bar} \times 12.23 \, \text{L} = 12.23 \, \text{L bar}
\]
Converting bar to atm:
\[
W = 12.23 \, \text{L bar} \times 0.987 \, \text{atm/bar} \approx 12.07 \, \text{L atm}
\]
### Final Answer
The work done by the gas is approximately **12 L atm**.
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