See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

f i T 4T  f f V i 1 T v S nC ln nRln T v    f 1 v 3R 0 n ln4 n.Rln 2 v    i f v 3log4 log 2 v  3/2 i f v 4 v  f i 8v v  f p i f T Pi 0 S nC ln nR.ln T P    5/2 f i P 4 P  i f 32P P 