Deux exos sur les cycles
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@ -2769,6 +2769,121 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{solos}
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\end{exos}
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\begin{exos}
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On fait passer \SI{32}{\gram} de méthane (\(CH_4\)) de l'état (\SI{1}{\bar} ; \SI{60}{\celsius}) à l'état (\SI{5}{\bar} ; \SI{60}{\celsius}) par une compression isotherme, puis, par une compression adiabatique, à l'état (\SI{30}{\bar} ; ?). Calculez our chaque transformations~:
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\begin{enumerate}
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\item le travail dépensé,
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\item la chaleur échangée avec le milieu extérieur et
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\item la variation d'énergie interne.
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\end{enumerate}
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\begin{solos}
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Comme la masse atomique du carbone vaut \SI{12}{uma} et celle de l'hydrogène \SI{1}{uma}, celle de la molécule de méthane (\(CH_4\)) vaut \SI{16}{uma}. Ainsi, sa masse molaire vaut \SI{16}{\gram\per\mol}. Si on a \SI{32}{\gram} de ce gaz, on a donc \SI{2}{\mol} de \(CH_4\).
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\subsubsection*{Compression isotherme}
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Avec une température de 273 + 60 = \SI{333}{\kelvin}, on peut alors calculer les volumes~:
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\begin{align*}
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V_1&=\frac{n\cdot R\cdot T}{p_1}=\frac{2\cdot 8,31\cdot 333}{10^5}=\SI{0,055}{\metre\cubed}\\
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V_2&=\frac{n\cdot R\cdot T}{p_2}=\frac{2\cdot 8,31\cdot 333}{5\cdot 10^5}=\SI{0,011}{\metre\cubed}
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\end{align*}
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Le travail isotherme se calcule alors aisément par~:
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\begin{align*}
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A&=n\cdot R\cdot T\cdot ln(\frac{V_2}{V_1})\\
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&=2\cdot 8,31\cdot 333\cdot ln(\frac{0,011}{0,055}=\SI{-8907}{\joule}
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\end{align*}
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Comme la compression est isotherme, on a aussi~:
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\[\Delta U = 0\;\text{et}\;Q=A=\SI{-8907}{\joule}\]
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\subsubsection*{Compression adiabatique}
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La compression adiabatique se fait de l'état 2 à l'état 3. Comme vu précédemment, le volume de l'état 2 est \(V_2=\SI{0,011}{\metre\cubed}\).
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Des propriétés de la transformation adiabatique, on tire alors~:
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\begin{align*}
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p_2\cdot V_2&=p_3\cdot V_3\;\Rightarrow\; 5\cdot 0,011^{4/3}=30\cdot V_3^{4/3}\\
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V_3^{4/3}&=4,1\cdot 10^{-4}\;\Rightarrow\;V_3=\SI{2,87e-3}{\metre\cubed}
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\end{align*}
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Par la loi des gaz parfaits, on en tire que~:
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\[T_3=\frac{p_3\cdot V_3}{n\cdot R}=\SI{518}{\kelvin}=\SI{245}{\celsius}\]
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Le travail est alors~:
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\begin{align*}
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A&=-\frac{i}{2}\cdot (p_3\cdot V_3-p_2\cdot V_2)\\
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&=-\frac{6}{2}\cdot (30\cdot 10^5\cdot 2,87\cdot 10^{-3}-5\cdot 10^5\cdot 0,011)\\
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&=\SI{-9330}{\joule}
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\end{align*}
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Avec pour échange de chaleur et variation d'énergie interne~:
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\[Q=0\;\text{et}\;\Delta U=-A=\SI{9330}{\joule}\]
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\end{solos}
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\end{exos}
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\begin{exos}
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Une machine thermique cyclique travaille avec un gaz parfait monoatomique qui subit quatre transformations. Le tableau suivant présente les échange d'énergie au cours du cycle.
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\begin{center}
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\begin{tabular}{|c|c|c|c|}
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\hline
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Transformation & \(\Delta U\) & \(A\) & \(Q\) \\
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& J & J & J \\\hline
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1 & 0 & -1109 & ? \\
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2 & 7200 & ? & 0 \\
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3 & ? & 4436 & 4436 \\
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4 & ? & 7200 & ? \\
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\hline
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\end{tabular}
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\end{center}
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\smallskip
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\begin{enumerate}
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\item Complétez le tableau en justifiant vos résultats.
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\item Calculez le rendement.
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\item Le gaz se trouvant initialement dans l'état : \(V_0=\SI{16}{\deci\metre\cubed}\), \(p_0=\SI{1e5}{\pascal}\) et \(T_0=\SI{400}{\kelvin}\), calculez le produit \(n\cdot R\).
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\item Déterminez les températures à la fin des étapes 1, 2 et 3.
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\item Faites un diagramme de bilan du cycle.
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\end{enumerate}
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\begin{solos}
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Le tableau complété est le suivant :
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\begin{center}
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\begin{tabular}{|c|c|c|c|}
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\hline
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Transformation & \(\Delta U\) & \(A\) & \(Q\) \\
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& J & J & J \\\hline
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isotherme & 0 & -1109 & -1109 \\
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adiabatique & 7200 & -7200 & 0 \\
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isotherme & 0 & 4436 & 4436 \\
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adiabatique & -7200 & 7200 & 0 \\
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\hline
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\(\sum\) & 0 & 3327 & \\
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\hline
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\end{tabular}
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\end{center}
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\smallskip
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\begin{enumerate}
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\item Les trois premières lignes sont justifiées par le premier principe. Pour la dernière, on a utilisé le fait que la somme des variations des énergies internes sur un cycle fermé est nulle, puisqu'on se retrouve dans l'état initial. Ainsi :
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\[\sum \Delta U = 0 + 7200 + 0 -7200=0\]
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\item le rendement est donné par :
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\[\eta = \frac{\sum A}{\sum Q_+}=\frac{3327}{4436}=75\%\]
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\item La loi des gaz parfaits donne :
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\[n\cdot R=\frac{p_0\cdot V_0}{T_0}=\frac{10^5\cdot 16\cdot 10^{-3}}{400}=4\]
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\item La température à l'état initial \(T_0=\SI{400}{\kelvin}\) est donnée. La première transformation étant isotherme, on a que \(T_1=\SI{400}{\kelvin}\). Pour la seconde transformation, on peut écrire :
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\begin{align*}
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\Delta U&=\frac{i}{2}\cdot n\cdot R\cdot \Delta T\\
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7200&=\frac{3}{2}\cdot 4\cdot \Delta T\\
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\Delta T&=1200=T_2-T_1=T_2-400\\
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&\Rightarrow\;T_2=\SI{1600}{\kelvin}
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\end{align*}
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Et comme la troisième est isotherme, on a : \(T_3=\SI{1600}{\kelvin}\). \item Le diagramme de bilan est donné à la figure \ref{exos:cycle1}.
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\end{enumerate}
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\begin{figure}
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\def\svgwidth{7cm}
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\begin{center}
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%\input{Annexe-Exercices/Images/cycle2.eps_tex}
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\includegraphics[scale=0.9]{cycle2.eps}
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\end{center}
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\caption{Bilan du cycle\label{exos:cycle1}}
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\end{figure}
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\end{solos}
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\end{exos}
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%\begin{exos}
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% Un énoncé de test.
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% \begin{solos}
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@ -2769,6 +2769,120 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{solos}
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\end{exos}
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\begin{exos}
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On fait passer \SI{32}{\gram} de méthane (\(CH_4\)) de l'état (\SI{1}{\bar} ; \SI{60}{\celsius}) à l'état (\SI{5}{\bar} ; \SI{60}{\celsius}) par une compression isotherme, puis, par une compression adiabatique, à l'état (\SI{30}{\bar} ; ?). Calculez our chaque transformations~:
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\begin{enumerate}
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\item le travail dépensé,
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\item la chaleur échangée avec le milieu extérieur et
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\item la variation d'énergie interne.
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\end{enumerate}
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\begin{solos}
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Comme la masse atomique du carbone vaut \SI{12}{uma} et celle de l'hydrogène \SI{1}{uma}, celle de la molécule de méthane (\(CH_4\)) vaut \SI{16}{uma}. Ainsi, sa masse molaire vaut \SI{16}{\gram\per\mol}. Si on a \SI{32}{\gram} de ce gaz, on a donc \SI{2}{\mol} de \(CH_4\).
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\subsubsection*{Compression isotherme}
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Avec une température de 273 + 60 = \SI{333}{\kelvin}, on peut alors calculer les volumes~:
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\begin{align*}
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V_1&=\frac{n\cdot R\cdot T}{p_1}=\frac{2\cdot 8,31\cdot 333}{10^5}=\SI{0,055}{\metre\cubed}\\
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V_2&=\frac{n\cdot R\cdot T}{p_2}=\frac{2\cdot 8,31\cdot 333}{5\cdot 10^5}=\SI{0,011}{\metre\cubed}
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\end{align*}
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Le travail isotherme se calcule alors aisément par~:
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\begin{align*}
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A&=n\cdot R\cdot T\cdot ln(\frac{V_2}{V_1})\\
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&=2\cdot 8,31\cdot 333\cdot ln(\frac{0,011}{0,055}=\SI{-8907}{\joule}
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\end{align*}
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Comme la compression est isotherme, on a aussi~:
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\[\Delta U = 0\;\text{et}\;Q=A=\SI{-8907}{\joule}\]
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\subsubsection*{Compression adiabatique}
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La compression adiabatique se fait de l'état 2 à l'état 3. Comme vu précédemment, le volume de l'état 2 est \(V_2=\SI{0,011}{\metre\cubed}\).
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Des propriétés de la transformation adiabatique, on tire alors~:
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\begin{align*}
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p_2\cdot V_2&=p_3\cdot V_3\;\Rightarrow\; 5\cdot 0,011^{4/3}=30\cdot V_3^{4/3}\\
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V_3^{4/3}&=4,1\cdot 10^{-4}\;\Rightarrow\;V_3=\SI{2,87e-3}{\metre\cubed}
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\end{align*}
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Par la loi des gaz parfaits, on en tire que~:
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\[T_3=\frac{p_3\cdot V_3}{n\cdot R}=\SI{518}{\kelvin}=\SI{245}{\celsius}\]
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Le travail est alors~:
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\begin{align*}
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A&=-\frac{i}{2}\cdot (p_3\cdot V_3-p_2\cdot V_2)\\
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&=-\frac{6}{2}\cdot (30\cdot 10^5\cdot 2,87\cdot 10^{-3}-5\cdot 10^5\cdot 0,011)\\
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&=\SI{-9330}{\joule}
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\end{align*}
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Avec pour échange de chaleur et variation d'énergie interne~:
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\[Q=0\;\text{et}\;\Delta U=-A=\SI{9330}{\joule}\]
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\end{solos}
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\end{exos}
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\begin{exos}
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Une machine thermique cyclique travaille avec un gaz parfait monoatomique qui subit quatre transformations. Le tableau suivant présente les échange d'énergie au cours du cycle.
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\begin{center}
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\begin{tabular}{|c|c|c|c|}
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\hline
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Transformation & \(\Delta U\) & \(A\) & \(Q\) \\
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& J & J & J \\\hline
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1 & 0 & -1109 & ? \\
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2 & 7200 & ? & 0 \\
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3 & ? & 4436 & 4436 \\
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4 & ? & 7200 & ? \\
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\hline
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\end{tabular}
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\end{center}
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\smallskip
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\begin{enumerate}
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\item Complétez le tableau en justifiant vos résultats.
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\item Calculez le rendement.
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\item Le gaz se trouvant initialement dans l'état : \(V_0=\SI{16}{\deci\metre\cubed}\), \(p_0=\SI{1e5}{\pascal}\) et \(T_0=\SI{400}{\kelvin}\), calculez le produit \(n\cdot R\).
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\item Déterminez les températures à la fin des étapes 1, 2 et 3.
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\item Faites un diagramme de bilan du cycle.
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\end{enumerate}
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\begin{solos}
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Le tableau complété est le suivant :
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\begin{center}
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\begin{tabular}{|c|c|c|c|}
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\hline
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Transformation & \(\Delta U\) & \(A\) & \(Q\) \\
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& J & J & J \\\hline
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isotherme & 0 & -1109 & -1109 \\
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adiabatique & 7200 & -7200 & 0 \\
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isotherme & 0 & 4436 & 4436 \\
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adiabatique & -7200 & 7200 & 0 \\
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\hline
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\(\sum\) & 0 & 3327 & \\
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\hline
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\end{tabular}
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\end{center}
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\smallskip
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\begin{enumerate}
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\item Les trois premières lignes sont justifiées par le premier principe. Pour la dernière, on a utilisé le fait que la somme des variations des énergies internes sur un cycle fermé est nulle, puisqu'on se retrouve dans l'état initial. Ainsi :
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\[\sum \Delta U = 0 + 7200 + 0 -7200=0\]
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\item le rendement est donné par :
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\[\eta = \frac{\sum A}{\sum Q_+}=\frac{3327}{4436}=75\%\]
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\item La loi des gaz parfaits donne :
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\[n\cdot R=\frac{p_0\cdot V_0}{T_0}=\frac{10^5\cdot 16\cdot 10^{-3}}{400}=4\]
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\item La température à l'état initial \(T_0=\SI{400}{\kelvin}\) est donnée. La première transformation étant isotherme, on a que \(T_1=\SI{400}{\kelvin}\). Pour la seconde transformation, on peut écrire :
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\begin{align*}
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\Delta U&=\frac{i}{2}\cdot n\cdot R\cdot \Delta T\\
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7200&=\frac{3}{2}\cdot 4\cdot \Delta T\\
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\Delta T&=1200=T_2-T_1=T_2-400\\
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&\Rightarrow\;T_2=\SI{1600}{\kelvin}
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\end{align*}
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Et comme la troisième est isotherme, on a : \(T_3=\SI{1600}{\kelvin}\). \item Le diagramme de bilan est donné à la figure \ref{exos:cycle1}.
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\end{enumerate}
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\begin{figure}
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\def\svgwidth{7cm}
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\begin{center}
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\input{Annexe-Exercices/Images/cycle2.eps_tex}
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\end{center}
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\caption{Bilan du cycle\label{exos:cycle1}}
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\end{figure}
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\end{solos}
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\end{exos}
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%\begin{exos}
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% Un énoncé de test.
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% \begin{solos}
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281
Annexe-Exercices/Images/cycle1.eps
Normal file
281
Annexe-Exercices/Images/cycle1.eps
Normal file
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63
Annexe-Exercices/Images/cycle2.eps_tex
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Binary file not shown.
Binary file not shown.
@ -1055,6 +1055,88 @@
|
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|
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\end{Solution OS}
|
||||
\begin{Solution OS}{35}
|
||||
Comme la masse atomique du carbone vaut \SI{12}{uma} et celle de l'hydrogène \SI{1}{uma}, celle de la molécule de méthane (\(CH_4\)) vaut \SI{16}{uma}. Ainsi, sa masse molaire vaut \SI{16}{\gram\per\mol}. Si on a \SI{32}{\gram} de ce gaz, on a donc \SI{2}{\mol} de \(CH_4\).
|
||||
|
||||
\subsubsection*{Compression isotherme}
|
||||
Avec une température de 273 + 60 = \SI{333}{\kelvin}, on peut alors calculer les volumes~:
|
||||
\begin{align*}
|
||||
V_1&=\frac{n\cdot R\cdot T}{p_1}=\frac{2\cdot 8,31\cdot 333}{10^5}=\SI{0,055}{\metre\cubed}\\
|
||||
V_2&=\frac{n\cdot R\cdot T}{p_2}=\frac{2\cdot 8,31\cdot 333}{5\cdot 10^5}=\SI{0,011}{\metre\cubed}
|
||||
\end{align*}
|
||||
Le travail isotherme se calcule alors aisément par~:
|
||||
\begin{align*}
|
||||
A&=n\cdot R\cdot T\cdot ln(\frac{V_2}{V_1})\\
|
||||
&=2\cdot 8,31\cdot 333\cdot ln(\frac{0,011}{0,055}=\SI{-8907}{\joule}
|
||||
\end{align*}
|
||||
Comme la compression est isotherme, on a aussi~:
|
||||
\[\Delta U = 0\;\text{et}\;Q=A=\SI{-8907}{\joule}\]
|
||||
|
||||
\subsubsection*{Compression adiabatique}
|
||||
La compression adiabatique se fait de l'état 2 à l'état 3. Comme vu précédemment, le volume de l'état 2 est \(V_2=\SI{0,011}{\metre\cubed}\).
|
||||
|
||||
Des propriétés de la transformation adiabatique, on tire alors~:
|
||||
\begin{align*}
|
||||
p_2\cdot V_2&=p_3\cdot V_3\;\Rightarrow\; 5\cdot 0,011^{4/3}=30\cdot V_3^{4/3}\\
|
||||
V_3^{4/3}&=4,1\cdot 10^{-4}\;\Rightarrow\;V_3=\SI{2,87e-3}{\metre\cubed}
|
||||
\end{align*}
|
||||
Par la loi des gaz parfaits, on en tire que~:
|
||||
\[T_3=\frac{p_3\cdot V_3}{n\cdot R}=\SI{518}{\kelvin}=\SI{245}{\celsius}\]
|
||||
Le travail est alors~:
|
||||
\begin{align*}
|
||||
A&=-\frac{i}{2}\cdot (p_3\cdot V_3-p_2\cdot V_2)\\
|
||||
&=-\frac{6}{2}\cdot (30\cdot 10^5\cdot 2,87\cdot 10^{-3}-5\cdot 10^5\cdot 0,011)\\
|
||||
&=\SI{-9330}{\joule}
|
||||
\end{align*}
|
||||
Avec pour échange de chaleur et variation d'énergie interne~:
|
||||
\[Q=0\;\text{et}\;\Delta U=-A=\SI{9330}{\joule}\]
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{36}
|
||||
Le tableau complété est le suivant :
|
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\begin{center}
|
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\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Transformation & \(\Delta U\) & \(A\) & \(Q\) \\
|
||||
& J & J & J \\\hline
|
||||
isotherme & 0 & -1109 & -1109 \\
|
||||
adiabatique & 7200 & -7200 & 0 \\
|
||||
isotherme & 0 & 4436 & 4436 \\
|
||||
adiabatique & -7200 & 7200 & 0 \\
|
||||
\hline
|
||||
\(\sum\) & 0 & 3327 & \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\smallskip
|
||||
\begin{enumerate}
|
||||
\item Les trois premières lignes sont justifiées par le premier principe. Pour la dernière, on a utilisé le fait que la somme des variations des énergies internes sur un cycle fermé est nulle, puisqu'on se retrouve dans l'état initial. Ainsi :
|
||||
\[\sum \Delta U = 0 + 7200 + 0 -7200=0\]
|
||||
\item le rendement est donné par :
|
||||
\[\eta = \frac{\sum A}{\sum Q_+}=\frac{3327}{4436}=75\%\]
|
||||
\item La loi des gaz parfaits donne :
|
||||
\[n\cdot R=\frac{p_0\cdot V_0}{T_0}=\frac{10^5\cdot 16\cdot 10^{-3}}{400}=4\]
|
||||
\item La température à l'état initial \(T_0=\SI{400}{\kelvin}\) est donnée. La première transformation étant isotherme, on a que \(T_1=\SI{400}{\kelvin}\). Pour la seconde transformation, on peut écrire :
|
||||
\begin{align*}
|
||||
\Delta U&=\frac{i}{2}\cdot n\cdot R\cdot \Delta T\\
|
||||
7200&=\frac{3}{2}\cdot 4\cdot \Delta T\\
|
||||
\Delta T&=1200=T_2-T_1=T_2-400\\
|
||||
&\Rightarrow\;T_2=\SI{1600}{\kelvin}
|
||||
\end{align*}
|
||||
Et comme la troisième est isotherme, on a : \(T_3=\SI{1600}{\kelvin}\). \item Le diagramme de bilan est donné à la figure \ref{exos:cycle1}.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{figure}
|
||||
\def\svgwidth{7cm}
|
||||
\begin{center}
|
||||
%\input{Annexe-Exercices/Images/cycle2.eps_tex}
|
||||
\includegraphics[scale=0.9]{cycle2.eps}
|
||||
\end{center}
|
||||
\caption{Bilan du cycle\label{exos:cycle1}}
|
||||
\end{figure}
|
||||
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{37}
|
||||
Procédons simplement au calcul de l'incertitude absolue.
|
||||
|
||||
\smallskip
|
||||
@ -1102,7 +1184,7 @@
|
||||
Cette expression est légèrement différente de la précédente. Mais, le second terme est négligeable en raison de la présence de l'incertitude sur le temps au carré. On voit ainsi qu'il est nécessaire de faire attention aux ordres de grandeurs.
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{36}
|
||||
\begin{Solution OS}{38}
|
||||
\dots
|
||||
|
||||
\end{Solution OS}
|
||||
|
||||
1064
SolutionsOS.tex.bak
1064
SolutionsOS.tex.bak
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Reference in New Issue
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