Match the thermodynamic processes given under Column I with the expressionsgiven in Column IIExpansion of 1 mol of an ideal gas into avacuum under isolated conditionsMixing of equal volumes of two ideal gases atconstant temperature and pressure in an isolatedcontainerReversible heating of at 1 atm from 300K to 600 K, followed by reversible cooling to300 K at 1 atm
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answer is A-R;T,B-P;Q;S,C-P;Q;S,D-P;Q;S;T.
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Detailed Solution
A ® During freezing of water entropy decreases Δssys<0At equation conditions; DG = 0B® In vacuum, free expansion work done = 0 For an ideal gas in isolated condition (q = 0), DU = 0C® For an ideal gas and isolated condition, q = 0, DU = 0 As per first lawD® The process is cyclic through unique reversible pathHence DU = 0 (cyclic)DG = 0W = 0 (positive and negative work are same and opposite)And q = 0