One mole of N2O4 (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 N2O4(g) decomposes to NO2(g) . The resultant pressure is

Question

One mole of N2O4 (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 N2O4(g) decomposes to NO2(g) . The resultant pressure is (A)  1.2 atm  (B)  2.4 atm  (C)  2.0 atm  (D)  1.0 atm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ce3504f48df4db615e9fa6c86544c490-.png" /> Mole of unreacted N2O4 = 0.8 n1 = 1 n2 = 1.2 From pV = nRT (p1/T1n1) = (p2/T2n2) (1/300× 1) = (p2/600× 12) ⇒ p2 = (600× 1.2/300) = 2.4 atm

Q. One mole of $N_{2} O_{4} (g)$ at $300 K$ is kept in a closed vessel at $1 a t m$ pressure. It is heated to $600 K$ when $20%$ by mass of $N_{2} O_{4} (g)$ decomposes to $N O_{2} (g)$ . The resultant pressure is

$1.2 a t m$

$2.4 a t m$

$2.0 a t m$

$1.0 a t m$

Mole of unreacted $N_{2} O_{4} = 0.8$
$n_{1} = 1$
$n_{2} = 1.2$
From $p V = n RT$
$\frac{p _{1}}{T _{1} n _{1}} = \frac{p _{2}}{T _{2} n _{2}}$
$\frac{1}{300 \times 1} = \frac{p _{2}}{600 \times 12}$
$\Rightarrow p_{2} = \frac{600 \times 1.2}{300}$
$= 2.4 a t m$

Q. One mole of N2​O4​(g) at 300K is kept in a closed vessel at 1atm pressure. It is heated to 600K when 20% by mass of N2​O4​(g) decomposes to NO2​(g) . The resultant pressure is

1.2atm

2.4atm

2.0atm

1.0atm