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Question

Assuming that water vapor is an ideal gas, the internal energy change E when 1mol of water is vaporized at 1bar pressure and 100°C, (Given: molar enthalpy of vaporization of water at 1bar and 373K =41kJmol1 and R=8.3Jmol1K1 ) will be


A

4.100kJmol-1

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B

3.7904kJmol-1

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C

37.904kJmol-1

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D

41.00kJmol-1

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Solution

The correct option is C

37.904kJmol-1


Explanation for correct option

(C) 37.904kJmol-1

Internal energy

  • Internal energy is equal to the sum of internal kinetic energy and internal potential energy caused by molecular attraction forces.
  • Formula to calculate internal energy change is as follows:

E=H-ngRT

  • E is the internal energy change.
  • H is the enthalpy change.
  • R is a gas constant.
  • T is the temperature.
  • ng is the number of moles of gas species.

Determine the internal energy change

  • The vaporization of water is as follows:

H2OlH2Og

  • Enthalpy of vaporization of water is 41kJmol1.
  • R=8.3Jmol1K1
  • Temperature, T=373K
  • Number of moles, ng=1mol
  • the internal energy change is as follows:

E=H-ngRT=41kJmol-1-1mol×8.3Jmol-1K-1×10-3kJ1J×373K=41-3.096kJmol-1=37.904kJmol-1

Explanation for incorrect options

  • The internal energy change is 37.904kJmol-1.
  • Hence all the other options (A), (B) and (D) are incorrect.

Therefore, the correct option is (C) 37.904kJmol-1.


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