Match the thermodynamic processes given under Column I with expressions given in Column II

Explanation

`B - p, q, s C - p, q, s D - p, q, s, t`
During freezing of water entropy decrease, `Delta S _( sys) lt 0, Delta G = 0.` In vacum, free expansion work done `= 0 So, q = 0 and Delta U =0`
For an ideal gas in isolated condition `q =0 , Delta U = 0,` so as per first law `w =0.`
The process is cycolic through unique reversible path, hence `Delta H = 0 ` (cyclic), `Delta G =0.`
w = 0 (positive and negative work are same and oppostie ) and q =0.