See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

An adiabatic process implies q = 0, and with volume not changing we have V 0   ; therefore, W = 0 and thus U 0   . The constant ‘U’ and ‘V’ allows us to apply the strict entropy spontaneity test; if S 0   the process is spontaneous. Since, we are given that S is +ve, so the process must be spontaneous.    G H T S Here, H U 0,    S ve  So, G  must be negative.