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CGP EDU Academic Team
Published on: August 14, 2026
Water is boiled under a pressure of 1.0 atm. When an electric current of 0.50 amp from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.8 gm of water is vapourised. Determine the Δ E in kJ/mol in nearest possible integers at the boiling point of water. Given : R = 8.314 JK –1 mol –1 .
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
Sol. The enthalpy change for 0.8 gm of water
= Δ H ′ (say)
∴ Δ H ′ = 0.5 × 12 × 300 = 1800 J
Per mole Δ H = 
= 
= 40.5 kJ/mol
Δ E = Δ H – RT = 40.5 – 8.314 × 10 –3 × 373
= 40.5 – 3.1 = 34.4 kJ/mole
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