See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(for Q. 33 to 34) 0 0 0 0 A B A B P P 1 P P 2 2 2      0 0 A B P 3P 1atm 4 4   0 0 A B P 3P 4atm   and 0 0 0 C A B 4P P 3P 1 8 8 8    0 0 0 A B C P 3P 4P    = 8 atm.  0 0 A B P 3P  = (8 – 4 × 0.8) atm = 4.8 atm.