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Mathematics of Dipole Moment

At 300 K and ...

Question

At 300 K and one atmospheric pressure 2 L of $N_{2} O_{4} (g)$ was heated to 600 K at constant pressure. $50 %$ of $N_{2} O_{4} (g)$ got dissociated according to the reaction. $N_{2} O_{4} (g) ⇌ N_{2} (g) + O_{2} (g)$ Final volume of the mixture in litres is.

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Solution

$\begin{matrix} N_{2} O_{4} (g) & ⇋ & N_{2} (g) & + & 2 O_{2} (g) at 300 K: & a mole & 0 & 0 after 50 % : & a (1 - 0.5) & 0.5 a & 1.0 a dissociation \end{matrix}$ $\begin{matrix} P_{1} = 1 atm & P_{2} = 1 atm V_{1} = 2 L & V_{2} = ? n_{1} = a mole & n_{2} = 2 a moles T_{1} = 300 K & T_{2} = 600 K \end{matrix}$ $∵ \frac{P_{1} V_{1}}{n_{1} T_{1}} = \frac{P_{2} V_{2}}{n_{2} T_{2}}$ $∴ \frac{2}{300 a} = \frac{V_{2}}{2 \times 600 a}$

$V_{2} = 8 L$

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