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A car airbag is inflated by $\mathrm{N}_2$ produced during rapid decomposition of $\mathrm{NaN}_3(s)$. What will be the pressure of the inflated airbag if $130 \mathrm{~g}$ of $\mathrm{NaN}_3$ is used. The volume of airbag is $10 \mathrm{~L}\left(T=300 \mathrm{~K}, R=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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Verified Answer
The correct answer is:
$7.38 \mathrm{~atm}$
Given that,
$2 \mathrm{NaN}_3 \rightleftharpoons 2 \mathrm{Na}+3 \mathrm{~N}_2$
Number of moles of sodium azide $=\frac{130}{65}=2$
Number of moles of nitrogen $=2 \times \frac{3}{2}=3$
We know that, $V=\frac{n R T}{p}$
$\therefore \quad 10=\frac{3 \times 0.0821 \times 300}{p}$
$\begin{aligned} \Rightarrow \quad p & =\frac{3 \times 0.0821 \times 300}{10}=7.389 \mathrm{~atm} \\ & \simeq 7.38 \mathrm{~atm}\end{aligned}$
$2 \mathrm{NaN}_3 \rightleftharpoons 2 \mathrm{Na}+3 \mathrm{~N}_2$
Number of moles of sodium azide $=\frac{130}{65}=2$
Number of moles of nitrogen $=2 \times \frac{3}{2}=3$
We know that, $V=\frac{n R T}{p}$
$\therefore \quad 10=\frac{3 \times 0.0821 \times 300}{p}$
$\begin{aligned} \Rightarrow \quad p & =\frac{3 \times 0.0821 \times 300}{10}=7.389 \mathrm{~atm} \\ & \simeq 7.38 \mathrm{~atm}\end{aligned}$
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