5 mole of an ideal gas expand isothermally and irreversibly from a pressure of 10 atm to 1 atm against a constant external
pressure of 2 atm. `w_(irr)` at 300 K is :

Options

A. `-22.447KJ`
B. `-11.224KJ`
C. `-110.83KJ`
D. None of these

Explanation

To find the work done (w_irr) during the isothermal and irreversible expansion of an ideal gas, we can follow these steps:

### Step 1: Identify the Given Values
- Number of moles (n) = 5 moles
- Initial pressure (P1) = 10 atm
- Final pressure (P2) = 1 atm
- External pressure (P_external) = 2 atm
- Temperature (T) = 300 K
- Universal gas constant (R) = 8.314 J/(K·mol)

### Step 2: Write the Formula for Work Done in Irreversible Expansion
The work done during an irreversible expansion against a constant external pressure is given by:
\[ w_{irr} = -P_{external} \Delta V \]
Where \( \Delta V = V_2 - V_1 \).

### Step 3: Express Volume in Terms of Pressure
Using the ideal gas law, we can express volume (V) as:
\[ V = \frac{nRT}{P} \]
Thus, we can write:
\[ \Delta V = V_2 - V_1 = \frac{nRT}{P_2} - \frac{nRT}{P_1} \]

### Step 4: Substitute \(\Delta V\) into the Work Formula
Substituting \(\Delta V\) into the work formula gives:
\[ w_{irr} = -P_{external} \left( \frac{nRT}{P_2} - \frac{nRT}{P_1} \right) \]
This can be simplified to:
\[ w_{irr} = -P_{external} \cdot nRT \left( \frac{1}{P_2} - \frac{1}{P_1} \right) \]

### Step 5: Plug in the Values
Now, substituting the known values into the equation:
\[ w_{irr} = -2 \cdot 5 \cdot 8.314 \cdot 300 \left( \frac{1}{1} - \frac{1}{10} \right) \]
Calculating the term in the parentheses:
\[ \frac{1}{1} - \frac{1}{10} = 1 - 0.1 = 0.9 \]
Thus, we have:
\[ w_{irr} = -2 \cdot 5 \cdot 8.314 \cdot 300 \cdot 0.9 \]

### Step 6: Calculate the Work Done
Now calculating the values:
\[ w_{irr} = -2 \cdot 5 \cdot 8.314 \cdot 300 \cdot 0.9 \]
\[ = -2 \cdot 5 \cdot 8.314 \cdot 270 \]
\[ = -2 \cdot 5 \cdot 2244.78 \]
\[ = -22447.8 \text{ Joules} \]
Converting to kilojoules:
\[ w_{irr} = -22.4478 \text{ kJ} \]

### Final Answer
The work done \( w_{irr} \) at 300 K is approximately:
\[ w_{irr} \approx -22.45 \text{ kJ} \]