JEE Advanced (Matrix Match & Integer Answer):Thermodynamics | Chapter-wise Tests for JEE Main & Advanced PDF Download

DIRECTION : Question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q.1. Match the transformations in column I with appropriate options in column II (2011)
 Column-I                                               Column-II
 (A) CO₂ (s) → CO₂ (g)                         (p) phase transition
 (B) CaCO₃ (s) → CaO(s)+ CO₂ (g)     (q) allotropic change
 (C) 2Hg → H₂ (g)                                  (r) ΔH is positive
 (D) P_{(white, solid)} → P_{(red, solid)}      (s) ΔS is positive (t) ΔS is negative

Ans. Sol. A – p, r, s ; B – r, s ; C – t ; D – p, q, t
(A) CO₂ (s) → CO₂ (g) It is phase transition. The process is endothermic (sublimation). Gas is produced, so entropy increases.
(B) On heating CaCO₃ decomposes. So, process is endothermic.
The entropy increases as gaseous product is formed.
(C) 2H• → H2(g) Entropy decreases as number of gaseous particles decreases.
(D) It is phase transition.
White and red P are allotropes.
Red P is more stable than white.
So ΔH is –ve.

DIRECTION : Question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

Q.2. Match the thermodynamic processes given under Column-I with the expressions given under Column-II.         (JEE Adv. 2015)
 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      (r) ΔS_sys < 0 in an isolated container
 (D) Reversible heating of H₂(g) at 1 atm from 300 K to 600 K, followed by reversible          (s) ΔU = 0 cooling to 300 K at 1 atm
                                                                                                                                                         (t) ΔG = 0

Ans. Sol.  A- (r, t); B -(p, q, s); C -(p, q, s); D- (p, q, s, t) (A) → r,t 

It is at equilibrium at 273 K and 1 atm So ΔS_sys is negative As it is equilibrium process so ΔG = 0

(B) → p, q, s

Expansion of 1 mole of an ideal gas in vacuum under isolated condition
Hence, w = 0 and q_p = C_pdT  (∵dT = 0) ⇒ q = 0
ΔU = C_vdT        (∵dT = 0)    ΔU = 0
(C) → p, q, s 
Mixing of two ideal gases at constant temperature
Hence, DT = 0
∴ q = 0;  ΔU = 0 also w = 0      
(ΔU = q + w)
(D) → p, q, s, t 
Reversible heating and cooling of gas follows same path also initial and final position is same.