If the boundary of system moves by an infinitesimal amount, the work involved is given by `dw=-P_("ext")dV`
for irreversible process `w=-P_("ext")DeltaV " "( "where "DeltaV=V_(f)-V_(i))`
for reversible process `P_("ext")=P_("int")pmdP~=P_("int")`
so for reversible isothermal process `w = -nRTln.(V_(f))/(V_(i))`
2mole of an ideal gas undergoes isothermal compression along three different plaths :
(i) reversible compression from `P_(i)=2` bar and `V_(i) = 8L` to `P_(f) = 20` bar
(ii) a single stage compression against a constant external pressure of 20 bar, and
(iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until `P_("gas")=P_("ext")`, followed by compression against a constant pressure of 20 bar until `P_("gas") = P_("ext")`
Work done on the gas in single stage compression is :

Options

A. 36
B. 72
C. 144
D. None of these

Explanation

To solve the problem of work done on the gas during a single-stage compression against a constant external pressure of 20 bar, we will follow these steps:

### Step 1: Understand the Work Done Formula
For an irreversible process where the gas is compressed against a constant external pressure, the work done (w) can be expressed as:
\[ w = -P_{\text{ext}} \Delta V \]
where \( \Delta V = V_f - V_i \).

### Step 2: Identify Initial and Final Conditions
Given:
- Initial Pressure \( P_i = 2 \) bar
- Initial Volume \( V_i = 8 \) L
- Final Pressure \( P_f = 20 \) bar
- External Pressure \( P_{\text{ext}} = 20 \) bar

### Step 3: Calculate Initial Conditions
Using the ideal gas equation \( PV = nRT \):
- For the initial state:
\[ P_i V_i = nRT \implies 2 \, \text{bar} \times 8 \, \text{L} = 2 \, \text{moles} \times R \times T \]
Converting pressure from bar to atm (1 bar = 0.98692 atm):
\[ 2 \, \text{bar} \approx 1.972 \, \text{atm} \]
So,
\[ 1.972 \times 8 = 2RT \]
This gives us the product \( RT \).

### Step 4: Calculate Final Volume
Using the final pressure:
\[ P_f V_f = nRT \implies 20 \, \text{bar} \times V_f = 2RT \]
Converting \( P_f \) to atm:
\[ 20 \, \text{bar} \approx 19.44 \, \text{atm} \]
Thus,
\[ 19.44 \times V_f = 2RT \]

### Step 5: Relate Initial and Final Volumes
From the equations:
1. \( 1.972 \times 8 = 2RT \)
2. \( 19.44 \times V_f = 2RT \)

Equating the two:
\[ 1.972 \times 8 = 19.44 \times V_f \]
Solving for \( V_f \):
\[ V_f = \frac{1.972 \times 8}{19.44} \]

### Step 6: Calculate \( \Delta V \)
Now calculate \( \Delta V \):
\[ \Delta V = V_f - V_i \]
Substituting \( V_i = 8 \, \text{L} \) and the calculated \( V_f \).

### Step 7: Calculate Work Done
Now substitute \( \Delta V \) into the work done formula:
\[ w = -P_{\text{ext}} \Delta V \]
Substituting \( P_{\text{ext}} = 20 \, \text{bar} \) and the calculated \( \Delta V \).

### Step 8: Final Calculation
Calculate the final value of work done \( w \).

### Final Answer
The work done on the gas during the single-stage compression is calculated as per the above steps.