`5` moles of a liqiud `L` are converted into its vapour at its boiling point `( 273^(@)C)` and at a pressure of `1 "atm"`. If the value of latent heat of vapourisation of liquid `L` is `273 L atm//"mole"`, then which of the following statements is`//`are correct. Assume volume of liquid to be negligible and valpour of the liquid to behave ideally.

Options

A. Work done by the system in the above process is `224 L` atm.
B. The enthalpy change `(Delta H)` for the above process is `1365 L` atm (with respect to magnitude only)
C. The internal energy of the system increases in the above process.
D. The value of `Delta U` for the above process is `1589 L` atm.

Explanation

To solve the problem step by step, we will analyze the process of converting 5 moles of liquid L into vapor at its boiling point and at a pressure of 1 atm. We will calculate the work done, enthalpy change, and internal energy change.

### Step 1: Calculate the Volume of Vapor
Using the ideal gas law, we can calculate the volume of the vapor produced. The formula for volume \( V \) is given by:

\[ V = nRT \]

Where:
- \( n = 5 \) moles
- \( R = 0.0821 \, \text{L atm/(K mol)} \) (ideal gas constant)
- \( T = 273 + 273 = 546 \, \text{K} \) (boiling point in Kelvin)

Calculating the volume:

\[ V = 5 \times 0.0821 \times 546 = 224.0 \, \text{L} \]

### Step 2: Calculate Work Done by the System
The work done \( W \) by the system during the expansion can be calculated using the formula:

\[ W = -P_{\text{external}} \Delta V \]

Where:
- \( P_{\text{external}} = 1 \, \text{atm} \)
- \( \Delta V = V_{\text{final}} - V_{\text{initial}} = 224 \, \text{L} - 0 \, \text{L} = 224 \, \text{L} \)

Calculating the work done:

\[ W = -1 \times 224 = -224 \, \text{L atm} \]

### Step 3: Calculate Enthalpy Change (\( \Delta H \))
The enthalpy change can be calculated using the latent heat of vaporization:

\[ \Delta H = n \times L \]

Where:
- \( L = 273 \, \text{L atm/mole} \)

Calculating the enthalpy change:

\[ \Delta H = 5 \times 273 = 1365 \, \text{L atm} \]

### Step 4: Calculate Change in Internal Energy (\( \Delta U \))
The change in internal energy can be calculated using the relation:

\[ \Delta U = Q + W \]

Where:
- \( Q = \Delta H = 1365 \, \text{L atm} \)
- \( W = -224 \, \text{L atm} \)

Calculating the change in internal energy:

\[ \Delta U = 1365 - 224 = 1141 \, \text{L atm} \]

### Summary of Results
1. Work done by the system: \( -224 \, \text{L atm} \) (correct statement)
2. Enthalpy change: \( 1365 \, \text{L atm} \) (correct statement)
3. Internal energy increases (correct statement)
4. Change in internal energy: \( 1141 \, \text{L atm} \) (incorrect statement)

### Conclusion
The correct statements are:
- Work done by the system is \( 224 \, \text{L atm} \) (correct).
- Enthalpy change \( \Delta H \) is \( 1365 \, \text{L atm} \) (correct).
- Internal energy of the system increases (correct).
- The value of \( \Delta U \) is \( 1141 \, \text{L atm} \) (incorrect).