Q. 1 mole of $N_{2}O_{4}(g)$ at 300 K is kept in a closed container under 1 atm. 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

A) 1.2 atm
B) 2.4 atm
C) 2.0 atm
D) 1.0 atm
Explanation

(b)
$N_{2}O_{4} \rightleftharpoons 2NO_{2}$
Initial 1 mol 0
At eqm 1-0.20 0.40 mol
$pV = nRT$
$1 \times V = 1 \times R \times 300$ …(i)
$p \times V = 1.2 \times R \times 600$ …(ii)
On dividing (ii) by (i)
$p = 2.4$ atm

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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