Temperature and moles of gas given volume, pressure, enthalpy and constant pressure molar heat capacity
Hint: A general outline of what I think is the idea behind this problem/ This is a quick back of the envelope estimate: 
$$\ce{NaN3(s) -&gt; Na(s) + 3/2N2(g)}$$
The reaction given above takes place, and releases:

$$\Delta H = -23.1\ \mathrm{kJ} $$

60% of this heat is transferred to gaseous nitrogen, and since the process is adiabatic no additional heat enters/leaves the system. All of this energy goes into expansion work

$$Q = \frac{60}{100} \times 23.1 \ \mathrm{kJ} = 13.86\ \mathrm{kJ} $$

Temperature of $\ce{N2(g)}$ (assuming initially it is at $298\ \mathrm K$) can be found using (first, assuming one mole of azide decomposes releasing 3/2 moles of nitrogen)

$$Q = nC_{\mathrm m,p} \Delta T$$

$$T_{\text{final}} = \frac{Q}{nC_{\mathrm m,p}} + T_{\text{initial}} = 615.3\ \mathrm{K}$$

Volume occupied by 3/2 moles of nitrogen at the above temperature and given pressure is 

$$V = \frac{nRT}{P} = 30.3\ \mathrm{L}$$

You need another $9.7 \ \mathrm{L}$ of gas to completely fill the airbag, I am sure you can complete the calculation. I'll come back and post a full, detailed answer later.



Complete Answer

So, we need a volume of $ 40 \ \mathrm{L}$, at an unspecified temperature. 

$$\frac{PV}{n_{\ce{N2}}R} = T_\text{final}$$

Additionally, $$T_{\text{final}} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

I am assuming since the process is adiabatic no additionally heat enters/leaves the system. The way I see it, the energy generated during this decomposition would heat the nitrogen produced to some (as of yet unknown temperature), and would be used for doing expansion work. I assume everything is initially at $298 \ \mathrm{K}$

$$\frac{PV}{n_{\ce{N2}}R} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

From the reaction stoichiometry, $n_{\ce{N2}} = \frac{2}{3}n_{\ce{NaN3}}$

Thus, the expression we are interested in is

$$\frac{PV}{(2/3)n_{\ce{NaN3}}R} = \frac{Q}{(2/3)n_{\ce{NaN3}}C_{\mathrm m,p}} + T_{\text{initial}}$$

Now, it is just a matter of plugging in appropriate values, and solving for $n$ 

I used, 


$ P = 2.5 \ \mathrm{atm}$
$ V = 40 \  \mathrm{L}$
$ R = 0.0821 \ \mathrm{L \ atm \ mol^{-1} \ K^{-1}}$
$ Q = 0.6 \times 23.1 \times 1000 \times n_{\ce{NaN3}} \ \mathrm{kJ}$
$C_{\mathrm m,p} = 29.12 \ \mathrm{J\ mol^{−1}\ K^{−1}} $


Answer given below:


   Solving all of this I get no. of mol of azide $\approx 1.33401 \ \mathrm{mol}$ and, no. of mol of nitrogen produced $ \approx 2.001015 \ \mathrm{mol}$. Additionally, $T_{\text{final}} = 615.307 \ \mathrm{K}$ Plugging all of this back into the ideal gas equation and solving for volume at the given pressure gives me $40.4\ \mathrm{L}$