A sample containing 10 moles of a liquid (denoted as L) is undergoing 50% vaporization at its boiling temperature 273^∘ under conditions of constant external pressure of 1 atm. The value of the latent heat of vaporization of liquid L is 273 L atm/mol. Assume ideal behavior and constant external pressure throughout the vaporization. 

Match the following for the above case.

List I	List II
i) Work done by the system in the above process is	a) 1365 L atm
ii) The enthalpy change \(\Delta H\) for the above process is	b) 224 L atm
ii) The entropy of the system increases by	c) 1600 L atm
iv) The value of \(\Delta U\) for the above process is	d) 1589 L atm
v) The entropy of the system decreases by	e) 2.5 L atm

Thermodynamics of Vaporization

• The question involves thermodynamic calculations for converting 10 moles of a liquid (L) completely into vapor at its boiling point under constant pressure.
• During vaporization, enthalpy change (ΔH) is positive, as heat energy is absorbed by the system to overcome intermolecular forces.
• The latent heat provided (273 L atm/mol) represents energy required per mole to convert liquid into vapor at the boiling point.
• Thermodynamic relationships for enthalpy (ΔH), entropy (ΔS), internal energy (ΔU), and work done (W) are crucial to solving this problem.

Calculation:

Work done (W): The system expands during vaporization, performing work against external pressure. The volume change (ΔV) and pressure (P) are used to calculate this.

V1 = 5 × R × 546 = 224 L

Work done by the system = External pressure × Change in volume = 1 × 224 = 224 L atm

Therefore, Work done by the system = 224 L atm

Enthalpy change (ΔH): Calculated from the given latent heat and number of moles. It represents the heat absorbed during vaporization at constant pressure.

Enthalpy change (ΔH) = q = 273 × 5 = 1365 L atm

Internal energy change (ΔU): Obtained by subtracting the work done from the enthalpy change (ΔU = ΔH - W).

Internal energy change (ΔU) = q – W = 1365 − 224 = 1141 L atm