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 paths :
(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")`
Order of magnitude of work is :

Options

A. `w_(1) gt w_(2) gt w_(3)`
B. `w_(3) gt w_(2) gt w_(1)`
C. `w_(2) gt w_(3) gt w_(1)`
D. `w_(1)=w_(2)=w_(3)`

Explanation

To solve the problem of work done during isothermal compression of an ideal gas along three different paths, we will calculate the work done for each path step by step.

### Step 1: Calculate Work Done for Reversible Compression

For a reversible isothermal process, the work done \( W \) can be calculated using the formula:

\[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \]

Given:
- \( n = 2 \) moles
- \( P_i = 2 \) bar
- \( V_i = 8 \) L
- \( P_f = 20 \) bar

First, we need to find the initial and final volumes using the ideal gas law:

\[ PV = nRT \]

For initial conditions:

\[ V_i = \frac{nRT_i}{P_i} \]

Assuming isothermal conditions, we can use the same temperature for final conditions. We need to find \( V_f \) at \( P_f = 20 \) bar.

Using the ideal gas law for final conditions:

\[ V_f = \frac{nRT_f}{P_f} \]

Since \( T_i = T_f \), we can express \( V_f \) in terms of \( V_i \):

\[ \frac{V_f}{V_i} = \frac{P_i}{P_f} \]

Calculating \( V_f \):

\[ \frac{V_f}{8} = \frac{2}{20} \implies V_f = \frac{8 \times 2}{20} = 0.8 \, L \]

Now, substituting into the work formula:

\[ W = -2RT \ln \left( \frac{0.8}{8} \right) \]

### Step 2: Calculate Work Done for Single Stage Compression

For a single stage compression against a constant external pressure of \( P_{ext} = 20 \) bar, the work done is given by:

\[ W = -P_{ext} \Delta V \]

Where \( \Delta V = V_f - V_i = 0.8 - 8 = -7.2 \, L \).

Converting \( L \) to \( m^3 \) (1 L = 0.001 m³):

\[ \Delta V = -7.2 \times 0.001 = -0.0072 \, m^3 \]

Now substituting into the work formula:

\[ W = -20 \times 10^5 \times (-0.0072) = 14400 \, J \]

### Step 3: Calculate Work Done for Two Stage Compression

In the two-stage compression, we first compress against a constant pressure of \( 10 \) bar until the gas pressure equals the external pressure, and then compress against \( 20 \) bar.

1. **First Stage**: Compress against \( 10 \) bar until \( P_{gas} = 10 \) bar.
- Using the ideal gas law, we can find the volume at \( P = 10 \) bar.
- \( V = \frac{nRT}{10} \)
- Calculate \( V \) using the same \( T \) as before.

2. **Second Stage**: Compress against \( 20 \) bar from the volume found in the first stage to \( V_f \).

The total work done will be the sum of the work done in both stages.

### Summary of Work Done

1. **Reversible Compression**: \( W_1 \) (calculated from Step 1)
2. **Single Stage Compression**: \( W_2 = 14400 \, J \)
3. **Two Stage Compression**: \( W_3 \) (calculated from Step 3)

### Conclusion

The order of magnitude of work done is:

\[ W_2 > W_3 > W_1 \]