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Thermodynamics — Easy JEE chemistry MCQ

For vaporization of $1$ mole of water at $100°C$ and $1$ atm, if $\Delta H_{vap} = 40.7$ kJ/mol, the entropy change is:
  1. A. $109$ J/mol·K
  2. B. $40.7$ J/mol·K
  3. C. $407$ J/mol·K
  4. D. $10.9$ J/mol·K

Solution

$\Delta S = \frac{\Delta H}{T} = \frac{40700}{373} = 109$ J/mol·K

CHEMISTRY

easy • PYQ Reworded
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For vaporization of 11 mole of water at 100°C100°C and 11 atm, if ΔHvap=40.7\Delta H_{vap} = 40.7 kJ/mol, the entropy change is:
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Thermodynamics — Easy JEE Chemistry MCQ | MyGoalPrep