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Most Important Selected Qs for JEE AdvancedChemistryThermodynamics (C)

Paragraph A monoatomic ideal gas undergoes state change from initial state (A) to final state (E) through two paths. A B and D E are isothermal; B C and C D are adiabatic processes aligned & Work =|60| J from A B , work =|50| J from B C & Work =|40| J from C D , work =|35| J from D E aligned Question Select the correct statement about internal energy or work done for the above state change from A to E

Options

  1. A( S _ A E )_ Path-1 = ( S _ A E )_ Path-2 and H _ B c = H _ D E
  2. BS _ A , E = S _ A , B + S _ D , E and H _ A , B = H _ D , E =0
  3. CS _ A B = S _ D E and H _ B C = H _ A E
  4. DS _ B C = S _ C E and H _ A B = H _ C D

Correct answer

B. S _ A , E = S _ A , B + S _ D , E and H _ A , B = H _ D , E =0

Step-by-step solution

E_A _B and E_D _E are zero because T=0 ( E_ Path-1 =( E)_ Patt-2 = E_B c+ E_c 0 .B C and C D are adiabatic so E=w( E )_ Path -2 =-50+35=-15 ~J

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