# Temperature and moles of gas given volume, pressure, enthalpy and constant pressure molar heat capacity

I've been assigned this question as homework and have been attempting it for hours but cannot seem to get anywhere. Any help would be greatly appreciated!

An airbag of volume $40\ \mathrm L$ is inflated adiabatically through the following reaction:

$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$

One mole of $\ce{NaN3}$ decomposed produces an enthalpy of $-23.1\ \mathrm{kJ}$. $60\ \%$ of this energy is deposited in the gaseous product; assume the decomposition is instantaneous. $C_{\mathrm m,p}(\ce{N2})=29.12\ \ \mathrm{J\ mol^{-1}\ K^{-1}}$. When fully inflated, the pressure of the $\ce{N2}$ gas inside the airbag is $2.5\ \mathrm{atm}$.

Estimate the mass of $\ce{NaN3}$ needed to operate the airbag and state any assumptions or approximations being made.

So far, I'm only sure that I've done the following correctly:

$H=U+PV=q+w+PV$ where $q=0$ since the process is adiabatic, so $H=w+PV$

For the expansion of $\ce{N2}$ gas:

$H=0.6n(23.1\ \mathrm{kJ})=n(13.86\ \mathrm{kJ})$ where $n$ is moles of $\ce{NaN3}$

$n(13.86\ \mathrm{kJ})=U+PV$ so $U=n(13.86\ \mathrm{kJ})-(2.5\ \mathrm{atm})(40\ \mathrm L)=n(13.86\ \mathrm{kJ}-RT)$

$q=0$, $P=2.5\ \mathrm{atm}$, $V=40\ \mathrm L$ and $C_{\mathrm m,p}(\ce{N2})= 29.12\ \mathrm{J\ mol^{-1}\ K^{-1}}$

I'm not sure what else to do from here. I'd really appreciate any advice or if anyone could point me in the right direction. Thank you!

Hint: A general outline of what I think is the idea behind this problem/ This is a quick back of the envelope estimate: $$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$ 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.

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,

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

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 $40.4\ \mathrm{L}$