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 1 atm. `w_(irr)` at 300 K is :
(a)`-15.921 kJ`
(b)`-11.224 kJ`
(c)`-110.83 kJ`
(d)None of these

Options

A. `-15.921 kJ`
B. `-11.224 kJ`
C. `-110.83 kJ`
D. None of these

Explanation

To find the work done by an ideal gas during an isothermal and irreversible expansion, we can use the following steps:

### Step 1: Understand the Work Done in Irreversible Expansion
The work done (w) in an irreversible process can be expressed as:
\[ w = -P_{\text{external}} \Delta V \]
where \( P_{\text{external}} \) is the constant external pressure and \( \Delta V \) is the change in volume.

### Step 2: Calculate the Change in Volume
Using the ideal gas equation:
\[ PV = nRT \]
we can express volume (V) as:
\[ V = \frac{nRT}{P} \]
For the initial state (1) and final state (2):
- At \( P_1 = 10 \, \text{atm} \):
\[ V_1 = \frac{nRT}{P_1} = \frac{5 \times 8.314 \, \text{J/(mol K)} \times 300 \, \text{K}}{10 \, \text{atm}} \]
- At \( P_2 = 1 \, \text{atm} \):
\[ V_2 = \frac{nRT}{P_2} = \frac{5 \times 8.314 \, \text{J/(mol K)} \times 300 \, \text{K}}{1 \, \text{atm}} \]

### Step 3: Calculate \( V_1 \) and \( V_2 \)
1. Convert pressure from atm to J/m³ (1 atm = 101.325 J/L = 101325 J/m³):
- \( P_1 = 10 \, \text{atm} = 1013250 \, \text{J/m}^3 \)
- \( P_2 = 1 \, \text{atm} = 101325 \, \text{J/m}^3 \)

2. Calculate \( V_1 \):
\[ V_1 = \frac{5 \times 8.314 \times 300}{1013250} \approx 0.0123 \, \text{m}^3 \]

3. Calculate \( V_2 \):
\[ V_2 = \frac{5 \times 8.314 \times 300}{101325} \approx 0.123 \, \text{m}^3 \]

### Step 4: Calculate \( \Delta V \)
\[ \Delta V = V_2 - V_1 = 0.123 - 0.0123 = 0.1107 \, \text{m}^3 \]

### Step 5: Calculate Work Done \( w \)
Substituting \( \Delta V \) and \( P_{\text{external}} \):
\[ w = -P_{\text{external}} \Delta V = -101325 \, \text{J/m}^3 \times 0.1107 \, \text{m}^3 \]
\[ w \approx -11224 \, \text{J} = -11.224 \, \text{kJ} \]

### Conclusion
The work done \( w_{\text{irr}} \) at 300 K is:
**(b) -11.224 kJ**