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ExamsJEE AdvancedChemistry

A system of mass 100 kg undergoes a process in which its specific entropy increases from 0.3 kJ/(kg*K) to 0.4 kJ/(kg*K). At the same time, the entropy of the surroundings decreases from 80 kJ/K to 75 kJ/K. Find the total change in entropy of the universe, delta_S_universe, in kJ/K.

  1. 5
  2. 10
  3. 15
  4. 20

Correct answer: 5

Solution

Change in entropy of system: delta_S_sys = 100 kg * (0.4 - 0.3) kJ/(kg*K) = 100 * 0.1 = 10 kJ/K. Change in entropy of surroundings: delta_S_surr = 75 - 80 = -5 kJ/K. delta_S_universe = delta_S_sys + delta_S_surr = 10 + (-5) = 5 kJ/K.

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