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")`
Total work done on the gas in two stage compression is :

Options

A. 40
B. 80
C. 160
D. None of these

Explanation

To solve the problem of calculating the total work done on the gas during a two-stage compression process, we will follow these steps systematically:

### Step 1: Identify the Initial Conditions
We have 2 moles of an ideal gas undergoing isothermal compression. The initial conditions are:
- Initial Pressure, \( P_i = 2 \) bar
- Initial Volume, \( V_i = 8 \) L

### Step 2: Calculate Initial Work Done in the First Stage
In the first stage, the gas is compressed against a constant external pressure of \( P_{ext} = 10 \) bar until the gas pressure equals the external pressure.

Using the formula for work done during compression:
\[ w = -P_{ext} \Delta V \]
where \( \Delta V = V_f - V_i \).

To find \( V_f \) when \( P_{gas} = P_{ext} \):
Using the ideal gas law:
\[ PV = nRT \]
At the end of the first stage:
\[ P_{ext} \cdot V_f = nRT \]
Substituting \( P_{ext} = 10 \) bar, \( n = 2 \) moles, and \( R = 0.0831 \, L \cdot bar/(K \cdot mol) \) (assuming isothermal conditions at a constant temperature \( T \)):
\[ V_f = \frac{nRT}{P_{ext}} = \frac{2 \cdot 0.0831 \cdot T}{10} \]

Now, we need to find \( T \):
Using the initial conditions:
\[ P_i \cdot V_i = nRT \implies T = \frac{P_i \cdot V_i}{nR} = \frac{2 \cdot 8}{2 \cdot 0.0831} = \frac{16}{0.1662} \approx 96.3 \, K \]

Now substituting \( T \) back into the equation for \( V_f \):
\[ V_f = \frac{2 \cdot 0.0831 \cdot 96.3}{10} \approx 1.60 \, L \]

Now, calculate the work done in the first stage:
\[ \Delta V = V_f - V_i = 1.60 - 8 = -6.4 \, L \]
\[ w_1 = -10 \cdot (-6.4) = 64 \, L \cdot bar \]

### Step 3: Calculate Work Done in the Second Stage
In the second stage, the gas is compressed against a constant external pressure of \( P_{ext} = 20 \) bar until the gas pressure equals the external pressure.

Using the same formula:
\[ w_2 = -P_{ext} \Delta V \]
Now, we need to find the new volume \( V_f \) when \( P_{gas} = 20 \) bar:
\[ V_f = \frac{nRT}{P_{ext}} = \frac{2 \cdot 0.0831 \cdot 96.3}{20} \approx 0.83 \, L \]

Now calculate the work done in the second stage:
\[ \Delta V = V_f - V_f \text{(from first stage)} = 0.83 - 1.60 = -0.77 \, L \]
\[ w_2 = -20 \cdot (-0.77) = 15.4 \, L \cdot bar \]

### Step 4: Total Work Done
The total work done on the gas during the two-stage compression is:
\[ W_{total} = w_1 + w_2 = 64 + 15.4 = 79.4 \, L \cdot bar \]

### Final Answer
The total work done on the gas in the two-stage compression is approximately **79.4 L·bar**.