Subiectul II

Un cilindru orizontal, de volum V=41,55 \text{dm}^3, este împărţit în două compartimente printr-un piston uşor, termoizolant, ce se poate deplasa fără frecare. Primul compartiment conţine oxigen ( \mu_1=32 \text{g}/\text{mol} ) la temperatura​ t_1=15^{\circ}\textrm{C}, iar al doilea compartiment conţine argon \mu_2=40 \text{g}/\text{mol} ) la temperatura​ t_2=7^{\circ}\textrm{C}. Pistonul se află în echilibru mecanic. Masa oxigenului reprezintă 70\% din masa totală de gaz din cilindru, iar cantitatea totală de gaz din cilindru este \nu =4,7 \text{mol}. Determinați:

  1. masa oxigenului din primul compartiment;
  2. raportul \frac{V_1}{V_2} dintre volumele ocupate de cele două gaze;
  3. temperatura până la care ar trebui încălzit argonul, oxigenul păstrându-şi temperatura, astfel încât raportul dintre volumele ocupate de cele două gaze, când pistonul ajunge din nou în echilibru, să devină \frac{V_{1f}}{V_{2f}}=1,5 ;
  4. presiunea argonului din al doilea compartiment, în condiţiile specificate la subpunctul c.

Rezolvare:

  1. Oxigenul este gaz biatomic, iar argonul este gaz monoatomic:

\begin{align*} \Rightarrow \mu_{_{\text{O}_2}}&=2\cdot 16 \\& =32\ \frac{\text{g}}{\text{mol}} \end{align*}       

\begin{align*} \mu_{_{\text{Ar}}}=40\ \frac{\text{g}}{\text{mol}} \end{align*}

\begin{align*} \nu_{\text{total}}=\nu_{_{\text{O}_2}}+\nu_{_{\text{Ar}}} \end{align*}

\begin{align*} \nu=\frac{m}{\mu} \end{align*}

\begin{align*} m_{ _{O_2}}&=70\%\cdot m_{\text{total}} \\& =0,7\cdot m_{\text{total}} \end{align*}

\begin{align*} \Rightarrow m_{_{\text{Ar}}}&=30\% \cdot m_{\text{total}} \\& =0,3\cdot m_{\text{total}} \end{align*}

\begin{align*} \Rightarrow \nu_{_{\text{O}_2}}=\frac{m_{_{\text{O}_2}}}{\mu_{_{\text{O}_2}}} \end{align*}

\begin{align*}\Rightarrow \nu_{_{\text{Ar}}}=\frac{m_{_{\text{Ar}}}}{\mu_{_{\text{Ar}}}} \end{align*}

\begin{align*} \Rightarrow \nu_{\text{total}}&=\frac{m_{_{\text{O}_2}}}{\mu_{_{\text{O}_2}}}+\frac{m_{_{\text{Ar}}}}{\mu_{_{\text{Ar}}}} \\\\ & =\frac{0,7\cdot m_{\text{total}}}{\mu_{_{\text{O}_2}}}+\frac{0,3\cdot m_{_{\text{total}}}}{\mu_{_{\text{Ar}}}} \\\\ & = m_{\text{total}}\cdot \left (\frac{0,7}{\mu_{_{\text{O}_2}}}+\frac{0,3}{\mu_{_{\text{Ar}}}} \right ) \\\\ & =m_{\text{total}}\cdot\left ( \frac{0,7}{32} +\frac{0,3}{40}\right ) \\\\&=m_{\text{total}}\cdot \frac{0,7\cdot 40+0,3\cdot 32}{32\cdot 40} \\\\& =m_{\text{total}}\cdot\frac{37,6}{1280} \end{align*}

\begin{align*} \Rightarrow m_{\text{total}}&=\nu_{\text{total}}\cdot \frac{1280}{37,6} \\\\& =4,7\cdot \frac{1280}{37.6} \\\\& =160\ \text{g} \end{align*}

\begin{align*} m_{\text{O}_2}&=0,7\cdot m_{\text{total}} \\ &=0,7\cdot 160 \\ &=112\ \text{g} \end{align*}

\begin{align*} \Leftrightarrow m_{\text{O}_2}=112\ \text{g} \end{align*}.

  1. Pistonul se află în echilibru mecanic, deci presiunea în cele două compartimente este egală:

\Rightarrow p_{_{\text{O}_2}}=p_{_{\text{Ar}}}=p

p\cdot V=\nu\cdot R\cdot T

\Rightarrow p\cdot V_{_{\text{O}_2}}=\nu_{_{\text{O}_2}}\cdot R\cdot T_{_{\text{O}_2}}

\Rightarrow V_{_{\text{O}_2}}=\frac{\nu_{_{\text{O}_2}}\cdot R\cdot T_{_{\text{O}_2}}}{p}\quad(1)

p\cdot V_{_{\text{Ar}}}=\nu_{_{\text{Ar}}}\cdot R\cdot T_{_{\text{Ar}}}

\Rightarrow V_{_{\text{Ar}}}=\frac{\nu_{_{\text{Ar}}}\cdot R\cdot T_{_{\text{Ar}}}}{p}\quad(2)

T=t+273

\Rightarrow T_{_{\text{O}_2}}=15+273=288\ \text{K}

T_{_{\text{Ar}}}=7+273=280\ \text{K}

Împărţim relaţiile (1) şi (2) şi obţinem:

\begin{align*} \frac{V_{_{\text{O}_2}}}{V_{_{\text{Ar}}}}&=\frac{\dfrac{\nu_{_{\text{O}_2}}\cdot R\cdot T_{_{\text{O}_2}}}{p}}{\dfrac{\nu_{_{\text{Ar}}}\cdot R\cdot T_{_{\text{Ar}}}}{p}} \\\\& =\frac{\nu_{_{\text{O}_2}}\cdot T_{_{\text{O}_2}}}{\nu_{_{\text{Ar}}}\cdot T_{_{\text{Ar}}}} \\\\& =\frac{\dfrac{m_{_{\text{O}_2}}}{\mu_{_{\text{O}_2}}}\cdot T_{_{\text{O}_2}}}{\dfrac{m_{_{\text{Ar}}}}{\mu_{_{\text{Ar}}}}\cdot T_{_{\text{Ar}}}} \\\\& =\frac{\dfrac{m_{_{\text{O}_2}}}{\mu_{_{\text{O}_2}}}\cdot T_{_{\text{O}_2}}}{\dfrac{m_{_{\text{total}}}-m_{_{\text{O}_2}}}{\mu_{_{\text{Ar}}}}\cdot T_{_{\text{Ar}}}} \\\\& =\frac{\dfrac{112}{32}\cdot 288}{\dfrac{160-112}{40}\cdot 280} \\\\&=\frac{3,5\cdot 288}{\dfrac{48}{40}\cdot 280} \\\\& =\frac{1008}{1,2\cdot 280} \\\\& =\frac{1008}{336} \\\\& =3 \end{align*}

\begin{align*} \Leftrightarrow\frac{V_{_{\text{O}_2}}}{V_{_{\text{Ar}}}}& =3 \end{align*}.

  1. Avem

\frac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}=1,5 .

Folosim formula pentru raportul dintre volume obţinută la subpunctul anterior:

\begin{align*} \frac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}&=\frac{\dfrac{\nu_{_{\text{O}_2}}\cdot R\cdot T_{_{\text{O}_2}}}{p}}{\dfrac{\nu_{_{\text{Ar}}}\cdot R\cdot T'_{_{\text{Ar}}}}{p}} \\\\&=\frac{\nu_{_{\text{O}_2}}\cdot T_{_{\text{O}_2}}}{\nu_{_{\text{Ar}}}\cdot {T}'_{_{\text{Ar}}}} \end{align*}

\begin{align*} \Rightarrow T'_{_{\text{Ar}}}&=\frac{\nu_{_{\text{O}_2}}\cdot T_{_{\text{O}_2}}}{\nu_{_{\text{Ar}}}\cdot \dfrac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}} \\\\& =\frac{\dfrac{m_{\text{O}_2}}{\mu_{\text{O}_2}}\cdot T_{_{\text{O}_2}}}{\dfrac{m_{_{\text{Ar}}}}{\mu_{_{\text{Ar}}}}\cdot \dfrac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}} \\\\& =\frac{\dfrac{m_{_{\text{O}_2}}}{\mu_{_{\text{O}_2}}}\cdot T_{_{\text{O}_2}}}{\dfrac{m_{_{\text{total}}}-m_{_{\text{O}_{2}}}}{\mu_{_{\text{Ar}}}}\cdot \dfrac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}} \\\\& =\frac{\dfrac{112}{32}\cdot 288}{\dfrac{160-112}{40}\cdot 1,5} \\\\& =\frac{3,5\cdot 288}{1,2\cdot 1,5} \\\\& =\frac{1008}{1,8} \\\\& =560\ \text{K} \end{align*}

\begin{align*} \Leftrightarrow T'_{_{\text{Ar}}}&=560\ \text{K} \end{align*}.

  1. Avem

p\cdot V=\nu\cdot R\cdot T

\Rightarrow p'_{_{\text{Ar}}}\cdot V'_{_{\text{Ar}}}=\nu_{_{\text{Ar}}}\cdot R\cdot T'_{_{\text{Ar}}}

\begin{align*} \Rightarrow p'_{_{\text{Ar}}}&=\frac{\nu_{_{\text{Ar}}}\cdot R\cdot T'_{_{\text{Ar}}}}{V'_{_{\text{Ar}}}} \\\\& =\frac{\dfrac{m_{_{\text{Ar}}}}{\mu_{_{\text{Ar}}}}\cdot R\cdot T'_{_{\text{Ar}}}}{V'_{_{\text{Ar}}}} \end{align*}

\begin{align*} m_{_{\text{Ar}}}&=m_{_{\text{total}}}-m_{_{\text{O}_2}} \\&=160-112 \\&=48\ \text{g} \end{align*}

\begin{align*} \frac{V'_{_{\text{O}_2}}}{V'_{_{\text{Ar}}}}=1,5 \end{align*}

\begin{align*} \Rightarrow V'_{_{\text{O}_2}}=1,5\cdot V'_{_{\text{Ar}}} \end{align*}

Dar \begin{align*} V'_{_{\text{O}_2}}+V'_{_{\text{Ar}}}=V \end{align*}.

\begin{align*} \Rightarrow 1,5\cdot V'_{_{\text{Ar}}}+V'_{_{\text{Ar}}}=V \end{align*}

\begin{align*} \Leftrightarrow 2,5\cdot V'_{_{\text{Ar}}}=V \end{align*}

\begin{align*} \Rightarrow V'_{_{\text{Ar}}}&=\frac{V}{2,5} \\\\&=\frac{41,55}{2,5}\ \text{dm}^3 \\\\& =16,62\ \text{dm}^3 \\\\&=16,62\cdot 10^{-3}\ \text{m}^3 \end{align*}

\begin{align*} \Rightarrow p'_{_{\text{Ar}}}&=\frac{\dfrac{48}{40}\cdot 8,31\cdot 560}{16,62\cdot 10^{-3}} \\\\& =\frac{1,2\cdot 8,31\cdot 560}{16,62}\cdot 10^{3} \\\\&=\frac{5584,32}{16,62}\cdot 10^3 \\\\&=336\cdot 10^3\ \text{Pa} \\\\&=336\ \text{kPa} \end{align*}

\begin{align*} \Leftrightarrow p'_{_{\text{Ar}}}&=336\ \text{kPa} \end{align*}.