Question

One mole of $N_2O_4$ gas at 300 K is kept in a closed container at 1 atm. It is heated to 600 K when 20 percent by mass of $N_2O_4$ decomposes to $NO_2(g)$. The resultant pressure in the container would be

• A. 1.2 atm
• B. 2.4 atm
• C. 2 .0 atm
• D. 1.0 atm

Answer 1

Answer: (B)

2.4 atm

Answer 2

$N_2O_4----->2 NO_2$ $Initial \, 1 \,mol$ Ateqm.$\,(1-0.2)mol \,\,0.4\,mol$ $=0.8\,mol$ Total moles after dissociation, $n_2=0-8+0-4= 1.2$ Initial temperature = 300 K $P_1 V = n_1 RT_1 or 1 \times V = 1 \times R \times 300$ ...(i) Temperature after dissociation = 600 K. No. of moles after dissociation, $n_2 = 12$ $P_2 \times V = 1.2 \times R \times 600 ...$(ii) Dividing eqn. (ii) by eqn. (i) $p_2=\frac{1.2\times 600}{300}=2.4 atm$