When an electric current of 0.50 A from a 12 V supply is passed for 1805/6 sec through a resistance in thermal contact with water maintained at 1 atm and 373 K, it is found that 0.9 g of water is vapo

Question

When an electric current of 0.50 A from a 12 V supply is passed for 1805/6 sec through a resistance in thermal contact with water maintained at 1 atm and 373 K, it is found that 0.9 g of water is vaporized. The molar internal energy change (in kJ/mol) of water is (Take 8.314×373 = 3100)

Answer:

33.00

Solution:

q = VIt = 12×0.5×1805/6 = 1805 J for 0.05 mol, so Δ H_vap = 36100 J/mol. Δ U = Δ H - RT = 36100 - 3100 = 33000 J/mol = 33 kJ/mol. Ans = 0033.

TestHub · Accepted Answer

q = VIt = 12×0.5×1805/6 = 1805 J for 0.05 mol, so Δ H vap = 36100 J/mol. Δ U = Δ H - RT = 36100 - 3100 = 33000 J/mol = 33 kJ/mol. Ans = 0033.

Chemistry - Thermodynamics - I Question with Solution | TestHub

Doubts & Discussion