KTUYSBeta
Log inSign up

Thermodynamics · JEE Preparation Chemistry

For an isolated system, the change in internal energy (Delta U) and change in entropy (Delta S) respectively are:

Practice question from Thermodynamics (JEE Preparation Chemistry), with the correct answer and a step-by-step solution.

easy · MCQ · Q-B2B791

For an isolated system, the change in internal energy (ΔU\Delta U) and change in entropy (ΔS\Delta S) respectively are:

  1. A.ΔU=0,ΔS=0\Delta U = 0, \Delta S = 0
  2. B.ΔU>0,ΔS<0\Delta U > 0, \Delta S < 0
  3. C.ΔU=0,ΔS>0\Delta U = 0, \Delta S > 0Correct answer
  4. D.ΔU<0,ΔS=0\Delta U < 0, \Delta S = 0

Solution

An isolated system cannot exchange energy or matter with its surroundings, so ΔU=0\Delta U = 0. According to the second law of thermodynamics, any spontaneous process in an isolated system leads to an increase in entropy, hence ΔS>0\Delta S > 0.

More from Thermodynamics

All of Thermodynamics

Practise more Chemistry

Sign up free for adaptive practice on this chapter, with instant feedback and hints on every question.

Start practising free

Free to start · No credit card needed