After calculating delta H (in KJ) for heating 96.0g of NaCl (s) from 20 degrees

Question

After calculating delta H (in KJ) for heating 96.0g of NaCl (s) from 20 degrees Celsius to 76.0 Celsius by assuming Cp= 50.50 J mol^-1 K^-1. What is the delta U for the process above? After calculating delta H (in KJ) for heating 96.0g of NaCl (s) from 20 degrees Celsius to 76.0 Celsius by assuming Cp= 50.50 J mol^-1 K^-1. What is the delta U for the process above? After calculating delta H (in KJ) for heating 96.0g of NaCl (s) from 20 degrees Celsius to 76.0 Celsius by assuming Cp= 50.50 J mol^-1 K^-1. What is the delta U for the process above?

Explanation / Answer

delta U = n x Cv x delta T

Given that, Cp = 50.50 J/mol.K

delta T = 56 K

n = mass / molar mass = 96.0 / 58.5 = 1.64 mol

we know that, Cp - Cv = R => Cv = 50.50 - 8.314 = 42.186 J/mol.K

delta U = 1.64 mol x 42.186 J/mol.K x 56 K

=> delta U = 3874.36 J

=> delta U = 3.87 kJ <<<<<<<<<<-----------(ANSWER)

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