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 (in bar-L) on the gas in reversible isothermal compression is :

Options

A. 9.212
B. 36.848
C. 18.424
D. None of these

Explanation

To solve the problem of calculating the work done on the gas during reversible isothermal compression, we will follow these steps:

### Step 1: Identify the Given Values
- Initial Pressure, \( P_i = 2 \) bar
- Initial Volume, \( V_i = 8 \) L
- Final Pressure, \( P_f = 20 \) bar
- Number of moles, \( n = 2 \)

### Step 2: Use the Ideal Gas Law to Find Final Volume
For an ideal gas, the relationship between pressure and volume at constant temperature is given by:
\[ P_i V_i = P_f V_f \]
We can rearrange this to find \( V_f \):
\[ V_f = \frac{P_i V_i}{P_f} \]
Substituting the known values:
\[ V_f = \frac{2 \, \text{bar} \times 8 \, \text{L}}{20 \, \text{bar}} = \frac{16}{20} = 0.8 \, \text{L} \]

### Step 3: Calculate the Work Done Using the Formula for Reversible Isothermal Process
The work done on the gas during a reversible isothermal process is given by:
\[ w = -nRT \ln \left( \frac{V_f}{V_i} \right) \]
However, we need to express \( RT \) in terms of the initial conditions. We can use the ideal gas law to find \( T \):
\[ PV = nRT \implies T = \frac{P_i V_i}{nR} \]
Substituting the values:
\[ T = \frac{2 \, \text{bar} \times 8 \, \text{L}}{2 \times R} = \frac{16}{2R} = \frac{8}{R} \]

### Step 4: Substitute \( T \) Back into the Work Formula
Now we can substitute \( T \) into the work equation:
\[ w = -n \left( \frac{8}{R} \right) R \ln \left( \frac{0.8}{8} \right) \]
This simplifies to:
\[ w = -8 \ln \left( \frac{0.8}{8} \right) \]
Calculating \( \frac{0.8}{8} = 0.1 \):
\[ w = -8 \ln(0.1) \]

### Step 5: Calculate \( \ln(0.1) \)
Using the properties of logarithms:
\[ \ln(0.1) = \ln(10^{-1}) = -\ln(10) \]
Thus:
\[ w = -8 \times (-\ln(10)) = 8 \ln(10) \]

### Step 6: Substitute the Value of \( \ln(10) \)
Using \( \ln(10) \approx 2.303 \):
\[ w = 8 \times 2.303 = 18.424 \, \text{bar-L} \]

### Final Answer
The work done on the gas in reversible isothermal compression is approximately:
\[ \boxed{18.424 \, \text{bar-L}} \]