Question:
An airbag is inflated by \(N_2\) produced during rapid decomposition of \(NaN_3(s)\). What will be the pressure of the inflated airbag if 130 g of \(NaN_3\) is used? The volume of airbag is 10 L \((T=300\ K,\ R=0.082\ L\ atm\ K^{-1}\ mol^{-1})\)
An airbag is inflated by \(N_2\) produced during rapid decomposition of \(NaN_3(s)\). What will be the pressure of the inflated airbag if 130 g of \(NaN_3\) is used? The volume of airbag is 10 L \((T=300\ K,\ R=0.082\ L\ atm\ K^{-1}\ mol^{-1})\)
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For airbag problems, first calculate moles of gas produced from the balanced chemical equation and then apply
\[
PV=nRT.
\]
Updated On: Jun 18, 2026
- 7.38 atm
- 4.92 atm
- 3.0 atm
- 9.84 atm
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the decomposition reaction.
Sodium azide decomposes as \[ 2NaN_3(s)\rightarrow 2Na(s)+3N_2(g) \]
Step 2: Calculate moles of \(NaN_3\).
Molar mass of \(NaN_3\): \[ 23+3(14)=65\ \text{g mol}^{-1} \] Therefore, \[ n(NaN_3)=\frac{130}{65}=2 \] moles.
Step 3: Calculate moles of \(N_2\) produced.
From the balanced equation, \[ 2\ \text{mol}\ NaN_3 \rightarrow 3\ \text{mol}\ N_2 \] Hence, \[ n(N_2)=3 \] moles.
Step 4: Apply the ideal gas equation.
Using \[ PV=nRT \] \[ P=\frac{nRT}{V} \] Substituting values, \[ P=\frac{(3)(0.082)(300)}{10} \] \[ =\frac{73.8}{10} \] \[ =7.38\ \text{atm} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{P=7.38\ \text{atm}} \] Hence, option (1) is correct.
Sodium azide decomposes as \[ 2NaN_3(s)\rightarrow 2Na(s)+3N_2(g) \]
Step 2: Calculate moles of \(NaN_3\).
Molar mass of \(NaN_3\): \[ 23+3(14)=65\ \text{g mol}^{-1} \] Therefore, \[ n(NaN_3)=\frac{130}{65}=2 \] moles.
Step 3: Calculate moles of \(N_2\) produced.
From the balanced equation, \[ 2\ \text{mol}\ NaN_3 \rightarrow 3\ \text{mol}\ N_2 \] Hence, \[ n(N_2)=3 \] moles.
Step 4: Apply the ideal gas equation.
Using \[ PV=nRT \] \[ P=\frac{nRT}{V} \] Substituting values, \[ P=\frac{(3)(0.082)(300)}{10} \] \[ =\frac{73.8}{10} \] \[ =7.38\ \text{atm} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{P=7.38\ \text{atm}} \] Hence, option (1) is correct.
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