One mole of N2O4(g) at 300K is kept in a closed container under one atmosphere. It is heated to 600K when 20 % by mass of N2O4(g) decomposes to (NO)2(g) . The resultant pressure is Patm . Calculate the value of 10P ?

Question

One mole of N2O4(g) at 300K is kept in a closed container under one atmosphere. It is heated to 600K when 20 % by mass of N2O4(g) decomposes to (NO)2(g) . The resultant pressure is Patm . Calculate the value of 10P ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

underset1 - 0 . 2 underset1N2 O4leftharpoons underset0 . 4 underset02 NO2 Number of moles at 300 K = 1 No. of moles at 600 K = 0.8 + 0.4 = 1.2 V is constant (P2/P1)=(n2 T2/n1 T1) (P2/1)=(1.2 × 600/1 × 300)=2.4atm10P=24

Q. One mole of $N_{2} O_{4} (g)$ at $300 K$ is kept in a closed container under one atmosphere. It is heated to $600 K$ when $20%$ by mass of $N_{2} O_{4} (g)$ decomposes to $(NO)_{2} (g)$ . The resultant pressure is $P a t m$ . Calculate the value of $10 P$ ?

$1 - 0.2 1 N_{2} O_{4} ⇌ 0.4 0 2 N O_{2}$
Number of moles at 300 K = 1
No. of moles at 600 K = 0.8 + 0.4 = 1.2
V is constant
$\frac{P _{2}}{P _{1}} = \frac{n _{2} T _{2}}{n _{1} T _{1}}$
$\frac{P _{2}}{1} = \frac{1.2 \times 600}{1 \times 300} = 2.4 a t m 10 P = 24$

Q. One mole of N2​O4​(g) at 300K is kept in a closed container under one atmosphere. It is heated to 600K when 20% by mass of N2​O4​(g) decomposes to (NO)2​(g) . The resultant pressure is Patm . Calculate the value of 10P ?

1−0.21N2​O4​​​⇌0.402NO2​​​ Number of moles at 300 K = 1 No. of moles at 600 K = 0.8 + 0.4 = 1.2 V is constant P1​P2​​=n1​T1​n2​T2​​ 1P2​​=1×3001.2×600​=2.4atm10P=24

Q. One mole of $N_{2} O_{4} (g)$ at $300 K$ is kept in a closed container under one atmosphere. It is heated to $600 K$ when $20%$ by mass of $N_{2} O_{4} (g)$ decomposes to $(NO)_{2} (g)$ . The resultant pressure is $P a t m$ . Calculate the value of $10 P$ ?

$1 - 0.2 1 N_{2} O_{4} ⇌ 0.4 0 2 N O_{2}$
Number of moles at 300 K = 1
No. of moles at 600 K = 0.8 + 0.4 = 1.2
V is constant
$\frac{P _{2}}{P _{1}} = \frac{n _{2} T _{2}}{n _{1} T _{1}}$
$\frac{P _{2}}{1} = \frac{1.2 \times 600}{1 \times 300} = 2.4 a t m 10 P = 24$