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JEE Advanced2015ChemistryThermodynamics (C)Actual

Match the thermodynamic processes given under Column I with the expressions given under Column II. Column I Column II A. Freezing of water at 273 K and 1 atm P. q = 0 B. Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions Q. w = 0 C. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container R. Δ S s y s < 0 D. Reversible heating of H 2 ( g

Options

  1. Aa-r;b-s,t;c-r,s,t;d-s,t;
  2. Ba-q,r;b-q;c-q;d-s,t;
  3. Ca-r,t;b-p,q,s;c-p,q,s;d-p,q,s,t;
  4. Da-q,r,s;b-s,t;c-q,r,s;d-t;

Correct answer

C. a-r,t;b-p,q,s;c-p,q,s;d-p,q,s,t;

Step-by-step solution

A - H 2 O ( l ) → H 2 O ( s ) at 273 K. & 1 atm ∆ H = - v e = q ∆ S s y s < 0 , ∆ G = 0 w ≠ 0 (as water expands on freezing), ∆ U ≠ 0 B - Free expansion of ideal gas. q = 0 w = 0 ∆ U = 0 ∆ S s y s > 0 ∆ G < 0 C - Mixing of equal volume of ideal gases at constant pressure & temp in an isolated container q = 0, w = 0, ∆ U = 0 , ∆ S s y s > 0 , ∆ G < 0 D - H 2 g 300 K R e v e r s i b l e ⟶ H e a t i n g , 1 a t m 600 K R e v e r s i b l e ⟶ C o o l i n g , 1 a t m 300 K q = 0 , w = 0 , ∆ U = 0 , ∆ G = 0 , ∆ S s y s =

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