A current of 2.0 A from a 10 V supply is passed for 100 s through a resistor in thermal contact with a liquid X (46 g/mol) kept at its normal boiling point 351 K and 1 atm. 2.3 g of X vaporizes. Find

Question

A current of 2.0 A from a 10 V supply is passed for 100 s through a resistor in thermal contact with a liquid X (46 g/mol) kept at its normal boiling point 351 K and 1 atm. 2.3 g of X vaporizes. Find the molar internal energy change of vaporization (in kJ/mol, nearest integer). (Take RT = 2900 J/mol)

Answer:

37.00

Solution:

q = VIt = 10×2.0×100 = 2000 J for 2.3/46 = 0.05 mol. Δ H = 40000 J/mol. Δ U = Δ H - RT = 40000 - 2900 = 37100 J/mol ≈ 37 kJ/mol. Ans = 0037.

TestHub · Accepted Answer

q = VIt = 10×2.0×100 = 2000 J for 2.3/46 = 0.05 mol. Δ H = 40000 J/mol. Δ U = Δ H - RT = 40000 - 2900 = 37100 J/mol ≈ 37 kJ/mol. Ans = 0037.

Chemistry - Thermodynamics - I Question with Solution | TestHub

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