Question

What is ∆E when 2.0 mole of liquid water vaporises at 100˚C? The heat of vaporisation, ∆H vap. of water at 100˚C is 40.66 kJmol^-1.

Answer 1

 ∆H_vap = 40.66 kJ/mol 

∆H_gas = 2 × 40.66 kJ

PV = nRT

1 × V = 2× 0.0821 × 373 

V = 61.2466 L

∆PV = 1(∆V) = 1(V_g – V_L)

2 moles 

∴ Mass of H₂ O = 36g

d = 1 gm/cm³

∴ V = 36 cm³= 36×10^–3L