Quantities of 16 g of oxygen and 14 g of nitrogen are contained in a closed bottle. The pressure ins — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Initial moles: n($O_2$) = 16/32 = 0.5 mol, n($N_2$) = 14/28 = 0.5 mol [11]. Total moles = 1.0 mol [11]. After removing 8 g of $O_2$ (0.25 mol), remaining $O_2$ = 0.25 mol [11]. Total final moles = 0.25 + 0.5 = 0.75 mol [11]. Final pressure = 4 * (0.75/1.0) = 3 atm [11].