Chemistry - Thermodynamics - II Question with Solution | TestHub
Match Column I (Process) with Column II (Thermodynamic consequence)
Column I | Column II |
|---|---|
P) Reversible adiabatic expansion of an ideal gas | i) q = 0, Δ Ssys = 0, w < 0 |
Q) Reversible isothermal expansion of an ideal gas | ii) Δ U = 0, q = -w > 0 |
R) Adiabatic free expansion of an ideal gas | iii) q = 0, w = 0, Δ Ssys > 0 |
S) One complete Carnot cycle | iv) Δ U = Δ H = Δ Ssys = 0 for the working substance |
Options:
Answer:
Solution:
Reversible adiabatic: no heat, isentropic, and the gas does work on the surroundings (P – i). Isothermal ideal gas: Δ U = 0, so the heat absorbed equals the work done (Q – ii). Free expansion against a vacuum: w = 0, q = 0, and irreversibility raises the gas entropy (R – iii). Every state function of the working substance returns to its starting value after a full cycle (S – iv).
