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NEET CHEMISTRYThermodynamicsMedium

Question

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol130 \text{ kJ mol}^{-1} at 27C27^{\circ}\text{C}, the entropy change for the process would be:

A

1.0 J mol1 K11.0 \text{ J mol}^{-1} \text{ K}^{-1}

B

0.1 J mol1 K10.1 \text{ J mol}^{-1}\text{ K}^{-1}

C

100 J mol1 K1100 \text{ J mol}^{-1}\text{ K}^{-1}

D

10 J mol1 K110 \text{ J mol}^{-1}\text{ K}^{-1}

Step-by-Step Solution

For a phase transition process (such as the conversion of liquid water to steam) occurring at a constant temperature, the change in entropy (ΔS\Delta S) is related to the enthalpy change of the transition (ΔH\Delta H) and the absolute temperature (TT) by the equation: ΔS=ΔHT\Delta S = \frac{\Delta H}{T}

Given data: Enthalpy change, ΔH=30 kJ mol1=30000 J mol1\Delta H = 30 \text{ kJ mol}^{-1} = 30000 \text{ J mol}^{-1} Temperature, T=27C=27+273 K=300 KT = 27^{\circ}\text{C} = 27 + 273 \text{ K} = 300 \text{ K}

Substituting the values into the formula: ΔS=30000 J mol1300 K\Delta S = \frac{30000 \text{ J mol}^{-1}}{300 \text{ K}} ΔS=100 J mol1 K1\Delta S = 100 \text{ J mol}^{-1}\text{ K}^{-1} .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsenthalpychangetransitionliquidcirctextc

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