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  • Need clarity, kindly explain! - Thermodynamics - NEET

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^{-1} \ at \ 27^{o}C, the entropy change for the process would be

  • Option 1)

    10 J mol^{-1}K^{-1}

  • Option 2)

    1.0 J mol^{-1}K^{-1}

  • Option 3)

    0.1 J mol^{-1}K^{-1}

  • Option 4)

    100 J mol^{-1}K^{-1}

 

Answers (1)

best_answer

 

Entropy for phase transition at constant pressure -

\Delta S= \frac{\Delta H_{Transition}}{T}

- wherein

Transition \RightarrowFusion, Vaporisition, Sublimation

\Delta H\RightarrowEnthalpy

\Delta E\RightarrowInternal Energy

T\RightarrowTransitional temperature

 

 Entropy change = \frac{\Delta H}{T}

= \frac{30\times 10^{3} J/mdi-k}{300}

 = 100 J mol^{-1}K^{-1} 

 


Option 1)

10 J mol^{-1}K^{-1}

Incorrect

Option 2)

1.0 J mol^{-1}K^{-1}

Incorrect

Option 3)

0.1 J mol^{-1}K^{-1}

Incorrect

Option 4)

100 J mol^{-1}K^{-1}

Correct

Posted by

prateek

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