Match the thermodynamic processes given under Column I with the expressions given under Column II.
← JEE Advanced 2015 · Paper 1
ChemistryMatrix match+8 / -4Question 40 of 60
JEE Advanced 2015 · Paper 1, Chemistry question 40
Column I
- AFreezing of water at 273 K and 1 atm
- BExpansion of 1 mol of an ideal gas into a vacuum under isolated conditions
- CMixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container
- DReversible heating of at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm
Column II
- P
- Q
- R
- S
- T
| P | Q | R | S | T | |
|---|---|---|---|---|---|
| A | |||||
| B | |||||
| C | |||||
| D |
Answer and worked solution
Official answer: A → R, T; B → P, Q, S; C → P, Q, S; D → P, Q, S, T
- (A) Freezing at the normal freezing point is an equilibrium phase change: ΔG = 0 (T). Order increases, so ΔS_sys < 0 (R).
- (B) Free expansion of an ideal gas in isolation: q = 0, w = 0 and ΔU = 0 (P, Q, S).
- (C) Mixing ideal gases at constant T and P in isolation: q = 0, w = 0 and ΔU = 0 (P, Q, S).
- (D) Reversible heating then cooling back to the same state is a cycle. Net q and w are zero (equal and opposite at constant P), and ΔU = ΔG = 0 (P, Q, S, T).
Answer: A → R, T; B → P, Q, S; C → P, Q, S; D → P, Q, S, T