One mole of an ideal monoatomic gas expanded irreversibly in two stage expansion.
`{:("Sate-1",(8.0" bar",4.0" litre",300K)),("State-2",(2.0" bar",16" litre",300K)),("State-3",(1.0" bar",32" litre",300K)):}`
Total heat absorbed by the gas in the process is :

Options

A. 116 J
B. 40 J
C. 4000 J
D. None of these

Answer

To find the total heat absorbed by the gas during the two-stage expansion, we can follow these steps:

### Step 1: Identify the Initial and Final States
We have three states for the gas:
- State 1: \( P_1 = 8.0 \, \text{bar}, V_1 = 4.0 \, \text{L}, T_1 = 300 \, \text{K} \)
- State 2: \( P_2 = 2.0 \, \text{bar}, V_2 = 16.0 \, \text{L}, T_2 = 300 \, \text{K} \)
- State 3: \( P_3 = 1.0 \, \text{bar}, V_3 = 32.0 \, \text{L}, T_3 = 300 \, \text{K} \)

### Step 2: Calculate Work Done in Each Stage
The work done during an irreversible expansion 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.

#### Work Done from State 1 to State 2
- External Pressure \( P_{\text{ext}} = 2.0 \, \text{bar} \)
- Change in Volume \( \Delta V = V_2 - V_1 = 16.0 \, \text{L} - 4.0 \, \text{L} = 12.0 \, \text{L} \)

Calculating work:
\[ W_1 = -P_{\text{ext}} \Delta V = -2.0 \, \text{bar} \times 12.0 \, \text{L} = -24.0 \, \text{L bar} \]
Converting \( \text{L bar} \) to Joules (1 L bar = 100 J):
\[ W_1 = -2400 \, \text{J} \]

#### Work Done from State 2 to State 3
- External Pressure \( P_{\text{ext}} = 1.0 \, \text{bar} \)
- Change in Volume \( \Delta V = V_3 - V_2 = 32.0 \, \text{L} - 16.0 \, \text{L} = 16.0 \, \text{L} \)

Calculating work:
\[ W_2 = -P_{\text{ext}} \Delta V = -1.0 \, \text{bar} \times 16.0 \, \text{L} = -16.0 \, \text{L bar} \]
Converting \( \text{L bar} \) to Joules:
\[ W_2 = -1600 \, \text{J} \]

### Step 3: Calculate Total Work Done
Now, we can find the total work done in the entire process:
\[ W_{\text{total}} = W_1 + W_2 = -2400 \, \text{J} + (-1600 \, \text{J}) = -4000 \, \text{J} \]

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Where:
- \( \Delta U \) is the change in internal energy,
- \( Q \) is the heat absorbed by the system,
- \( W \) is the work done by the system (which is negative in our case since work is done on the gas).

For an ideal monoatomic gas, the change in internal energy can be calculated using:
\[ \Delta U = \frac{3}{2} n R \Delta T \]
Since the temperature remains constant (isothermal process), \( \Delta T = 0 \):
\[ \Delta U = 0 \]

Thus, the equation simplifies to:
\[ 0 = Q - (-4000 \, \text{J}) \]
\[ Q = 4000 \, \text{J} \]

### Final Answer
The total heat absorbed by the gas in the process is:
\[ \boxed{4000 \, \text{J}} \]