Solution

V = 350 mL = 0.350 L

$T = 20^{o}{C = (273+20) K}$

T = 293 K

$n = \frac{\text {mass}}{\text {Molecule mass}}$

Propane - ${CH_3 , CH_2 , CH_3}$

Molar mass = 44 g / mol

$n = \frac{3.2 { g }}{44 { g} { mol}^{-1}}$

Solution

$\mathrm {P} = \frac{\text {n R T }}{V}$

$P=\frac{\left(\frac{3.2 \mathrm {~ g}}{44\mathrm {~ g~ mole}^{-1}}\right)\times\left(0.82 \mathrm {~ L~ atm ~ K}^{-1} \mathrm {~ mol}^{-1}\right)\times 293 K}{0.350 \mathrm {~ L}}$

$P= \left [\quad\quad\right] \text{atm}$

What is the density in $\mathrm {g~ L}^{-1}$ of ammonia at $0 { } ^oC$ and 1 atm of the gas in a 1.000 L bulbs weights 0.672g at 23 $^{0}C$ and 733.4 mm $H_g$ pressure

Solution

$PV = n R T$

$n =\frac{m}{M}\quad PV = \frac{m}{M} \mathrm {RT}$

$P = \frac{1}M{}.\frac{m}{V}. RT$

$\frac{m}{V} \rightarrow \text {density}$

Solution

$P = \frac {\rho R T }{M}$

$\text {P = 1 atm}$

$T = 25^ oC = 298 K$

$V = 1.000~ L$

$m = 0.672\mathrm {~ g} /M_{NH_3} = 17\mathrm ~ g~ {mol}^{-1}$

$\mathrm {\rho = 733.4~ mm ~ Hg}$

$P\left\{\begin{array}{l}\text { bar } \\ \text { atm } \\ \text { mm Hg/torr}\end{array}\right.$

$760 \mathrm {~ mm ~ Hg} - 1 \text {~ atm}$

$733 \mathrm {~ m ~ Hg} - ? \text {~ atm}$

Solution

$\mathrm{NO}_2\left(T_h\right)(Initial) \xrightarrow[\text { cooled }]{ }\mathrm {NO}_2 + \mathrm {N}_2\mathrm{O}_4\left (T_e\right)(Funal)$

$\text {law T:} 2~ NO_2(g)\rightarrow N_2 ~ O_4(g)$

$\mathrm {n} = 15.2\mathrm {~ g ~ of} ~ NO_2\quad (n_{NO_2}^o)$

$\ M_{NO_2} = 46 \mathrm {~ g ~ mol}^{-1}$

$n^f =\ n_{NO_2}^f +2 \ n_{N_2O_4}^f = n^0=\ n_{NO_2}^0$

Solution

$P_{\text {total}} = 0.500 \text { atm }$

$P=\frac{n_{NO_2}^{f} R T}{V}+\frac{n_{N_2 O_4} R T}{V}$

$P= P_{NO_2} + P_{N_2O_4}$

$V = 10.0 ~ L / ~ T = 25^oC = 298 ~ K$

$R = 0.082 \text { L atm K }^{-1} \mathrm {~ mol}^{-1}$

$R= 0.500 \text {~ atm}=\left(\frac{RT}{V}\right)$

$\left\{n_{NO_2}^f + n_{N_2O_4}^f\right\}$

Automobile air bags are inflated by nitrogen gas generated by the rapid decomposition of sodium azide

$N_a N_3 (s) : 2 N_a N_3 (s) \rightarrow 2N_a (s) + 3N_2 (g)$

If an air bag has a volume of 36 L and is to be filled with nitrogen gas at 1.15 atm and $26^oC$ , how many grams of $N_a N_3$ , must be decomposed

Solution

$N_a N_3 (s) \rightarrow \text {high density small volume}$

$2 N_a ~ N_3 (s) \rightarrow 2 N_a (s) + 3 N_2 (g)$

$\text { V = 36 L}$

$\text {P = 1.15 atm}$

$\mathrm {T} = 26 ^{o}C \quad (n_{N_2})$

$\text {PV = n R T}$

Solution

$3 N_2 \rightarrow \text {2 moles of} ~ Na N_3$

$n_{N_2} \rightarrow ? ~ n_{Na N_3}$

$n_{Na N_3} =\frac{m_{Na N_3}}{M_{Na N_3}}$

$m_{Na N_3}$

$\text {Molar volume at} ~ 0^{o}C ~ (273K) \text {and 1 atm}$

$\quad\text {Molecule} \hspace {0.5cm}\text {Volume}$

$2 e^{-}~ ~ \left[\begin{array}{ll}\mathrm{H}_2 &\quad22.415 \\ \mathrm{H}_2 & \quad 22.428\end{array}\right.$

$10 e^{-} \left[\begin{array}{ll}\mathrm{NH}_3 & 22.077 \\ \mathrm{CH}_4 & 22.360 \\ \mathrm{Ne} & 22.428\end{array}\right.$

$14 e^{-}\left[\begin{array}{ll}\mathrm{C}_2 \mathrm{H}_2 & 22.274 \\ \mathrm{CO} & 22.405\end{array}\right.$

$\text {Molar volume close to 22.4 L but not equal}$

$Z = \frac{PV}{nRT}$

$\text {Z = 1 for an ideal gas}$