Subiectul II

O cantitate $\nu=1$ mol de oxigen, cu masa molară $\mu = 32$ g/mol , considerat gaz ideal, se află într-o stare având $p_1 = 2 \cdot 10^5$ Pa și $t_1 = 27^\circ C$ . Gazul este încălzit la presiune constantă astfel încât volumul ocupat de oxigen se dublează. Determinați:

masa unei molecule de oxigen;
numărul de molecule de oxigen;
volumul inițial ocupat de oxigen;
densitatea oxigenului în starea finală.

Rezolvare:

Oxigenul are masa molară $\mu_{0}=32\ \frac{\text{g}}{\text{mol}}$ .

Într-un mol sunt $N_{A}=6,023\cdot 10^{23}$ molecule.

$\mu=\frac{m}{\nu}$

$\Rightarrow m=\mu\cdot \nu$ .

Pentru $\nu=1\ \text{mol}$ avem:

$\begin{align*} m_{O_{2}}&=32\cdot 1 \\& =32\ g \end{align*}$

$\begin{align*} m_{O_{2}}=N_{A}\cdot m_\text{molecul\u a} \end{align*}$

$\begin{align*} m_{O_{2}}=\mu\cdot \nu \end{align*}$

$\begin{align*} \nu=1 \end{align*}$

$\begin{align*} \Rightarrow \mu\cdot \nu=N_{A}\cdot m_\text{molecul\u a} \end{align*}$

$\begin{align*} \Rightarrow m_\text{molecul\u a}&=\frac{\mu\cdot \nu}{N_{A}} \\\\& =\frac{32\cdot 1}{6,023\cdot 10^{23}} \\\\& \simeq \frac{32}{6}\cdot 10^{-23} \\\\& \simeq 5,33\cdot 10^{-23}\ g \\\\& =5,33\cdot 10^{-23}\cdot 10^{-3}\ \text{kg} \\\\& =5,33\cdot 10^{-26}\ \text{kg} \end{align*}$

$\begin{align*} \Leftrightarrow m_\text{molecul\u a}&=5,33\cdot 10^{-26}\ \text{kg} \end{align*}$ .

Într-un mol sunt $\begin{align*} N_{A}=6,023\cdot 10^{23}\ \end{align*}$ molecule.

$\begin{align*} \Rightarrow N_{O_{2}}=N_{A} \end{align*}$

$\begin{align*} \Rightarrow N_{O_{2}}=6,023\cdot 10^{23}\ \text{molecule O}_{2} \end{align*}$ .

Avem:

$\begin{align*} t_{1}=27^{\circ}\text{C} \end{align*}$

$\begin{align*} &\Rightarrow T_{1}=27+273\\ &\Leftrightarrow T_1=300\ \text{K} \end{align*}$

$\begin{align*} p\cdot V=\nu\cdot R\cdot T \end{align*}$

$\begin{align*} \Rightarrow V=\frac{\nu\cdot R\cdot T}{p} \end{align*}$

$\begin{align*} \Rightarrow V_{1}=\frac{\nu\cdot R\cdot T_{1}}{p_{1}} \end{align*}$

$\begin{align*} \nu=1\ \text{mol} \end{align*}$

$\begin{align*} \Rightarrow V_{1}&=\frac{1\cdot 8,31\cdot 300}{2\cdot 10^{5}} \\\\& =4,155\cdot 300\cdot 10^{-5} \\\\& =12,465\cdot 100\cdot 10^{-5} \\\\& =1246,5\cdot 10^{-5} \\\\& =12,465\cdot 10^{-3} \ \text{m}^{3} \\\\& =12,465\ \text{L }\end{align*}$

$\begin{align*} \Leftrightarrow V_{1}& =12,465\ \text{L} \end{align*}$

$\begin{align*} \Leftrightarrow V_{1}& \cong 12,5\ \text{L} \end{align*}$ .

Succesiv, avem:

$\begin{align*} V_{2}&=2\cdot V_{1} \\&=2\cdot 12,465\ \text{L}\\&=24,930\ \text{L}\\ & =24,93\cdot 10^{-3}\ \text{m}^{3} \end{align*}$

$\begin{align*} \Leftrightarrow V_{2}&=24,93\cdot 10^{-3}\ \text{m}^{3} \end{align*}$ .

$\begin{align*} m_{O_{2}}&=\mu_{O_{2}}\cdot \nu_{O_{2}}\\& =32\cdot 1=32\ \text{g}\\& =32\cdot 10^{-3}\ \text{kg} \end{align*}$

$\begin{align*} \Leftrightarrow m_{O_{2}}&=32\cdot 10^{-3}\ \text{kg} \end{align*}$ .

$\begin{align*} \rho&=\frac{m_{O_{2}}}{V_{2}} \\\\& =\frac{32\cdot 10^{-3}}{24,93\cdot 10^{-3}} \\\\&\simeq1,28\ \frac{\text{kg}}{\text{m}^{3}} \\\\&\simeq 1,3\ \frac{\text{kg}}{\text{m}^{3}} \end{align*}$

$\begin{align*} \Leftrightarrow \rho&\simeq 1,3\ \frac{\text{kg}}{\text{m}^{3}} \end{align*}$ .