Question

The enthalpy change for the evaporation of a liquid at its boiling point 127°C is +40.32 $kJmol^{-1}$. What is the value of internal energy change for the above process at 127°C? (R=8.3 $JK^{-1}mol^{-1}$)

• A. +37.0 $kJmol^{-1}$
• B. -37.0 $kJmol^{-1}$
• C. +43.64 $kJmol^{-1}$
• D. +43.0 $kJmol^{-1}$
• E. -43.0 $kJmol^{-1}$

Answer 1

The Correct Option is A

We are given:

• Enthalpy change (\( \Delta H \)) = +40.32 kJ/mol
• Boiling point = 127°C = 400 K
• R = 8.3 J/(mol·K) = 0.0083 kJ/(mol·K) 

Use the relation:

\( \Delta H = \Delta U + \Delta nRT \)

In evaporation, 1 mole of liquid becomes 1 mole of gas, so:

\( \Delta n = 1 \)

Now plug values:

\( \Delta U = \Delta H - \Delta nRT = 40.32 - (1)(0.0083)(400) \)

\( \Delta U = 40.32 - 3.32 = \mathbf{+37.0\ kJ/mol} \)

Correct Answer:

+37.0 kJ/mol