`2` moles of an ideal gas is compressed from (`1` bar, `2` L) to `2` bar isothermally. Calculate magnitude of minimum possible work in change (in joules ). (Given : `1` bar L = `100` J, ln`2=0.7`)

Answer

To calculate the magnitude of the minimum possible work done during the isothermal compression of an ideal gas, we can use the formula for work done in an isothermal process:

\[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) \]

or alternatively,

\[ W = -nRT \ln \left( \frac{P_1}{P_2} \right) \]

Where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = absolute temperature in Kelvin
- \( V_1 \) = initial volume
- \( V_2 \) = final volume
- \( P_1 \) = initial pressure
- \( P_2 \) = final pressure

### Step 1: Identify the given values
- Number of moles, \( n = 2 \, \text{moles} \)
- Initial pressure, \( P_1 = 1 \, \text{bar} = 100 \, \text{J/L} \)
- Final pressure, \( P_2 = 2 \, \text{bar} = 200 \, \text{J/L} \)
- Initial volume, \( V_1 = 2 \, \text{L} \)

### Step 2: Calculate the final volume \( V_2 \)
Using the ideal gas law for isothermal processes, we have:

\[ P_1 V_1 = P_2 V_2 \]

Rearranging gives:

\[ V_2 = \frac{P_1 V_1}{P_2} \]

Substituting the known values:

\[ V_2 = \frac{(1 \, \text{bar}) \times (2 \, \text{L})}{2 \, \text{bar}} = \frac{2}{2} = 1 \, \text{L} \]

### Step 3: Calculate the work done using the pressure ratio
Now we can calculate the work done using the pressure ratio:

\[ W = -nRT \ln \left( \frac{P_1}{P_2} \right) \]

Since we do not have the temperature \( T \), we can express \( nRT \) in terms of the initial conditions:

From the ideal gas law:

\[ P_1 V_1 = nRT \]

Thus,

\[ nRT = P_1 V_1 = (1 \, \text{bar}) \times (2 \, \text{L}) = 2 \, \text{bar L} \]

Converting \( 2 \, \text{bar L} \) to joules:

\[ 2 \, \text{bar L} = 2 \times 100 \, \text{J} = 200 \, \text{J} \]

### Step 4: Substitute values into the work formula
Now substituting into the work formula:

\[ W = -200 \ln \left( \frac{1}{2} \right) \]

Using the property of logarithms:

\[ \ln \left( \frac{1}{2} \right) = -\ln(2) \]

Thus:

\[ W = 200 \ln(2) \]

Given \( \ln(2) = 0.7 \):

\[ W = 200 \times 0.7 = 140 \, \text{J} \]

### Final Answer
The magnitude of the minimum possible work done during the isothermal compression is:

\[ \boxed{140 \, \text{J}} \]