Question

One mole of $ N_2O_{4} (g) $ at $ 300 K $ is kept in a closed vessel at $ 1 \, atm $ pressure. It is heated to $ 600 K $ when $ 20\% $ by mass of $ N_2O_4(g) $ decomposes to $ NO_2(g) $ . The resultant pressure is

• A. $ 1.2 \, atm $
• B. $ 2.4 \, atm $
• C. $ 2.0 \, atm $
• D. $ 1.0 \, atm $

Answer 1

Answer: (B)

Mole of unreacted $N_2O_4 = 0.8 $
$ n_1 = 1$
$ n_2 = 1.2$
From $pV = nRT$
$\frac{p_1}{T_1n_1} = \frac{p_2}{T_2n_2}$
$\frac{1}{300\times 1} = \frac{p_2}{600\times 12}$
$\Rightarrow p_2 = \frac{600\times 1.2}{300}$
$ = 2.4 \,atm$