To solve the problem of calculating the irreversible work done (w_irr) during the isothermal expansion of an ideal gas, we can follow these steps:
### Step 1: Understand the Given Data
- Number of moles (n) = 5 moles
- Initial pressure (P1) = 10 atm
- Final pressure (P2) = 1 atm
- External pressure (P_ext) = 1 atm
- Temperature (T) = 300 K
- Ideal gas constant (R) = 8.314 J/(mol·K)
### Step 2: Use the Formula for Irreversible Work
The formula for the work done during an irreversible expansion against a constant external pressure is given by:
\[
w_{irr} = -P_{ext} \times (V_2 - V_1)
\]
### Step 3: Calculate the Volumes (V1 and V2)
Using the ideal gas law, \( PV = nRT \), we can find the volumes at the initial and final pressures.
1. **Calculate V1 (Volume at P1)**:
\[
V_1 = \frac{nRT}{P_1} = \frac{5 \, \text{moles} \times 8.314 \, \text{J/(mol·K)} \times 300 \, \text{K}}{10 \, \text{atm}}
\]
Convert atm to Pa (1 atm = 101325 Pa):
\[
V_1 = \frac{5 \times 8.314 \times 300}{10 \times 101325} = \frac{12497.0}{101325} \approx 0.1233 \, \text{m}^3
\]
2. **Calculate V2 (Volume at P2)**:
\[
V_2 = \frac{nRT}{P_2} = \frac{5 \, \text{moles} \times 8.314 \, \text{J/(mol·K)} \times 300 \, \text{K}}{1 \, \text{atm}}
\]
\[
V_2 = \frac{5 \times 8.314 \times 300}{1 \times 101325} = \frac{12497.0}{101325} \approx 0.1233 \, \text{m}^3 \times 10 = 1.233 \, \text{m}^3
\]
### Step 4: Substitute the Volumes into the Work Formula
Now, substitute V1 and V2 into the work formula:
\[
w_{irr} = -P_{ext} \times (V_2 - V_1)
\]
Substituting the values:
\[
w_{irr} = -1 \, \text{atm} \times (1.233 \, \text{m}^3 - 0.1233 \, \text{m}^3)
\]
Convert atm to J (1 atm = 101325 J/m³):
\[
w_{irr} = -101325 \, \text{Pa} \times (1.233 - 0.1233) \, \text{m}^3
\]
\[
w_{irr} = -101325 \times 1.1097 \approx -112,224.0 \, \text{J} \approx -112.224 \, \text{kJ}
\]
### Final Answer
Thus, the irreversible work done \( w_{irr} \) is approximately:
\[
\boxed{-112.224 \, \text{kJ}}
\]
