$(a)\;2.92atm\qquad(b)\;1.92atm\qquad(c)\;3atm\qquad(d)\;1atm$

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$N_2\iff 2N$

Initial Moles of $N_2 = \large\frac{1.4}{28}$

Initial Mole of 2N = 0

Mole after dissociation of $N_2 = \large\frac{1.4}{28}\times \large\frac{70}{100}$

Moles after dissociation of 2N = $\large\frac{1.4}{28}\times \large\frac{2\times30}{100}$

Total Mole = $\large\frac{1.4}{28}\times\large\frac{70}{100}+\large\frac{1.4\times60}{100\times28}$

$=\large\frac{1.4}{28}\times[\large\frac{130}{100}]$

$P\times5 = \large\frac{1.4\times130}{28\times100}\times0.0821\times1800$

P = 1.92 atm

Hence answer is (b)

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