Match the process given in Column - I with the entropy change in Column - II
`{: ( "Column I", " Column II"),( "(a)Reversible adiabatic ideal gas compression.", " (p)"DeltaS_("surr") = 0 ) , ("(b)Reversible isothermal ideal gas expansion." , "(q)" DeltaS_("system") = 0),(" (c)Adiabatic free expansion " (p_(ext) = 0)" of an ideal gas" , "(r)" DeltaS_("surr") gt 0 ) , ( "(d)Irreversible isothermal ideal gas compression." , " (s)" DeltaS_("surr") lt 0 ):}`

Answer

To solve the problem of matching the processes in Column I with the corresponding entropy changes in Column II, we will analyze each process step by step.

### Step-by-Step Solution:

1. **Process (a): Reversible adiabatic ideal gas compression**
- In a reversible adiabatic process, there is no heat exchange with the surroundings (q = 0).
- Therefore, the change in entropy for the system (ΔS_system) is 0, and the change in entropy for the surroundings (ΔS_surr) is also 0.
- **Match:** (a) → (p) ΔS_surr = 0 and ΔS_system = 0.

2. **Process (b): Reversible isothermal ideal gas expansion**
- In an isothermal process, the temperature remains constant, and the gas expands.
- The change in entropy for the system (ΔS_system) can be calculated using the formula ΔS = nR ln(V2/V1), which is greater than 0 since V2 > V1.
- The change in entropy for the surroundings (ΔS_surr) is negative because heat is absorbed from the surroundings (q > 0), leading to ΔS_surr < 0.
- **Match:** (b) → (s) ΔS_surr < 0.

3. **Process (c): Adiabatic free expansion (p_ext = 0) of an ideal gas**
- In free expansion, there is no work done (W = 0) and no heat exchange (q = 0).
- The entropy change for the system (ΔS_system) is positive because the gas expands into a vacuum, leading to an increase in disorder.
- The change in entropy for the surroundings (ΔS_surr) remains 0 since there is no heat exchange.
- **Match:** (c) → (r) ΔS_surr > 0.

4. **Process (d): Irreversible isothermal ideal gas compression**
- In an irreversible isothermal process, the gas is compressed, and the volume decreases (V2 < V1).
- The change in entropy for the system (ΔS_system) is negative since the gas is compressed (ΔS = nR ln(V2/V1), which is less than 0).
- The change in entropy for the surroundings (ΔS_surr) is positive because work is done on the system, leading to heat being expelled to the surroundings.
- **Match:** (d) → (q) ΔS_system < 0.

### Final Matches:
- (a) → (p) ΔS_surr = 0
- (b) → (s) ΔS_surr < 0
- (c) → (r) ΔS_surr > 0
- (d) → (q) ΔS_system < 0