diff --git a/Annexe-Exercices/Annexe-Exercices.tex b/Annexe-Exercices/Annexe-Exercices.tex index a34c2df..b6101fc 100644 --- a/Annexe-Exercices/Annexe-Exercices.tex +++ b/Annexe-Exercices/Annexe-Exercices.tex @@ -2903,7 +2903,38 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r \item Trouvez la température maximale du gaz au cours de la compression de A à B. \end{enumerate} \begin{solos} - Un autre corrigé de test. + On commence par calculer le produit nR~: + \[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\] + \begin{enumerate} + \item Puis, on calcule la chaleur échangée lors de la transformation~: + \begin{align*} + Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\ + &=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\ + &\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\ + &=5250+7350+2100=\SI{14700}{\joule} + \end{align*} + \item La transformation est linéaire. On peut la multiplier par le volume~: + \begin{align*} + p&=-3\cdot 10^8\cdot (V-0,01)\\ + pV&=-3\cdot 10^8\cdot (V-0,01)\cdot V + \end{align*} + Comme le gaz est parfait~: + \begin{align*} + &pV=-3\cdot 10^8\cdot (V-0,01)\cdot V=n\cdot R\cdot T\\ + &\Rightarrow\;T(V)=-3\cdot 10^7 (V^2-0,01\cdot V) + \end{align*} + car nR vaut dix. + \item Pour trouver le maximum, il suffit alors de dériver et d'annuler~: + \begin{align*} + &\frac{dT}{dV}=-3\cdot 10^7 (2V-0,01)=0\\ + &\Rightarrow\;2\cdot V=0,01\;\Rightarrow\;V=\SI{0,005}{\metre\cubed}=\SI{5}{\deci\metre\cubed} + \end{align*} + La température maximale est alors~: + \begin{align*} + &T(0,005)=-3\cdot 10^7 (0,005^2-0,01\cdot 0,005)\\ + &=\SI{750}{\kelvin}\\ + \end{align*} + \end{enumerate} \end{solos} \end{exos} @@ -2914,6 +2945,13 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r % \end{solos} %\end{exos} +%\begin{exos} +% Un énoncé de test. +% \begin{solos} +% Un autre corrigé de test. +% \end{solos} +%\end{exos} + } \subsection{Relatifs aux incertitudes} diff --git a/Annexe-Exercices/Annexe-Exercices.tex.bak b/Annexe-Exercices/Annexe-Exercices.tex.bak index 4fd5ba0..b6101fc 100644 --- a/Annexe-Exercices/Annexe-Exercices.tex.bak +++ b/Annexe-Exercices/Annexe-Exercices.tex.bak @@ -2890,7 +2890,8 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r \centering \caption{Le diagramme PV.\label{lediagpv}} \centering - \input{Annexe-Exercices/Images/DiagPV.eps_tex} + \def\svgwidth{0.9\columnwidth} + \input{Annexe-Exercices/Images/DiagPV.ps_tex} \end{figure} \begin{enumerate} \item Calculez la chaleur reçue par le gaz pour passer de A à B. @@ -2902,7 +2903,38 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r \item Trouvez la température maximale du gaz au cours de la compression de A à B. \end{enumerate} \begin{solos} - Un autre corrigé de test. + On commence par calculer le produit nR~: + \[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\] + \begin{enumerate} + \item Puis, on calcule la chaleur échangée lors de la transformation~: + \begin{align*} + Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\ + &=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\ + &\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\ + &=5250+7350+2100=\SI{14700}{\joule} + \end{align*} + \item La transformation est linéaire. On peut la multiplier par le volume~: + \begin{align*} + p&=-3\cdot 10^8\cdot (V-0,01)\\ + pV&=-3\cdot 10^8\cdot (V-0,01)\cdot V + \end{align*} + Comme le gaz est parfait~: + \begin{align*} + &pV=-3\cdot 10^8\cdot (V-0,01)\cdot V=n\cdot R\cdot T\\ + &\Rightarrow\;T(V)=-3\cdot 10^7 (V^2-0,01\cdot V) + \end{align*} + car nR vaut dix. + \item Pour trouver le maximum, il suffit alors de dériver et d'annuler~: + \begin{align*} + &\frac{dT}{dV}=-3\cdot 10^7 (2V-0,01)=0\\ + &\Rightarrow\;2\cdot V=0,01\;\Rightarrow\;V=\SI{0,005}{\metre\cubed}=\SI{5}{\deci\metre\cubed} + \end{align*} + La température maximale est alors~: + \begin{align*} + &T(0,005)=-3\cdot 10^7 (0,005^2-0,01\cdot 0,005)\\ + &=\SI{750}{\kelvin}\\ + \end{align*} + \end{enumerate} \end{solos} \end{exos} @@ -2913,6 +2945,13 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r % \end{solos} %\end{exos} +%\begin{exos} +% Un énoncé de test. +% \begin{solos} +% Un autre corrigé de test. +% \end{solos} +%\end{exos} + } \subsection{Relatifs aux incertitudes} diff --git a/Annexe-Exercices/Images/DiagPV.ps_tex b/Annexe-Exercices/Images/DiagPV.ps_tex index 365798a..29d5028 100644 --- a/Annexe-Exercices/Images/DiagPV.ps_tex +++ b/Annexe-Exercices/Images/DiagPV.ps_tex @@ -1,4 +1,4 @@ -%% Creator: Inkscape 1.1.2 (0a00cf5339, 2022-02-04), www.inkscape.org +%% Creator: Inkscape 1.2.2 (b0a8486541, 2022-12-01), www.inkscape.org %% PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 %% Accompanies image file 'DiagPV.ps' (pdf, eps, ps) %% @@ -38,7 +38,7 @@ \newcommand*\fsize{\dimexpr\f@size pt\relax}% \newcommand*\lineheight[1]{\fontsize{\fsize}{#1\fsize}\selectfont}% \ifx\svgwidth\undefined% - \setlength{\unitlength}{274.42586846bp}% + \setlength{\unitlength}{275.228706bp}% \ifx\svgscale\undefined% \relax% \else% @@ -50,21 +50,21 @@ \global\let\svgwidth\undefined% \global\let\svgscale\undefined% \makeatother% - \begin{picture}(1,0.98504279)% + \begin{picture}(1,0.981988)% \lineheight{1}% 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-599,6 +603,7 @@ labeltitle labeltitleyear labeldateparts + pluralothers nohashothers nosortothers noroman @@ -1043,7 +1048,7 @@ given - + prefix family @@ -2344,72 +2349,48 @@ Bibliographies/BiblioCoursOS.bib - HNHR08 - JR00 - FC05 - SV06 - FB05 - SV06 - JL06 - SH03 - SH03 - EL99 - EL99 - GSJ05 - SH03 - BM85 - BM85 - BM85 - SH03 - OG04 - BS07 - GC88 - GC88 - LJ04 - HRaFL03 - OP06 - HRaFL03 - HRaFL03 - SV06 - HRaFL03 - SU03 - SU03 - AK05 - AS06 - AS02 - JJD21 - JL96 - GG92 - JR07 - GC88 - JR00 - GC88 - GC88 - BS07 - HNHR08 - JR00 - FC05 - SV06 - FB05 - JL06 - SH03 - EL99 - GSJ05 - BM85 - OG04 - BS07 - GC88 - LJ04 - HRaFL03 - OP06 - SU03 - AK05 - AS06 - AS02 - JJD21 - JL96 - GG92 - JR07 + HNHR08 + JR00 + FC05 + SV06 + FB05 + SV06 + JL06 + SH03 + SH03 + EL99 + EL99 + GSJ05 + SH03 + BM85 + BM85 + BM85 + SH03 + OG04 + BS07 + GC88 + GC88 + LJ04 + HRaFL03 + OP06 + HRaFL03 + HRaFL03 + SV06 + HRaFL03 + SU03 + SU03 + AK05 + AS06 + AS02 + JJD21 + JL96 + GG92 + JR07 + GC88 + JR00 + GC88 + GC88 + BS07 diff --git a/CoursMecaniqueOSDF.pdf b/CoursMecaniqueOSDF.pdf index 61aae3d..c7e00d2 100644 Binary files a/CoursMecaniqueOSDF.pdf and b/CoursMecaniqueOSDF.pdf differ diff --git a/SolutionsOS.tex b/SolutionsOS.tex index c4027cd..4494994 100644 --- a/SolutionsOS.tex +++ b/SolutionsOS.tex @@ -1137,7 +1137,38 @@ \end{Solution OS} \begin{Solution OS}{37} - Un autre corrigé de test. + On commence par calculer le produit nR~: + \[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\] + \begin{enumerate} + \item Puis, on calcule la chaleur échangée lors de la transformation~: + \begin{align*} + Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\ + &=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\ + &\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\ + &=5250+7350+2100=\SI{14700}{\joule} + \end{align*} + \item La transformation est linéaire. On peut la multiplier par le volume~: + \begin{align*} + p&=-3\cdot 10^8\cdot (V-0,01)\\ + pV&=-3\cdot 10^8\cdot (V-0,01)\cdot V + \end{align*} + Comme le gaz est parfait~: + \begin{align*} + &pV=-3\cdot 10^8\cdot (V-0,01)\cdot V=n\cdot R\cdot T\\ + &\Rightarrow\;T(V)=-3\cdot 10^7 (V^2-0,01\cdot V) + \end{align*} + car nR vaut dix. + \item Pour trouver le maximum, il suffit alors de dériver et d'annuler~: + \begin{align*} + &\frac{dT}{dV}=-3\cdot 10^7 (2V-0,01)=0\\ + &\Rightarrow\;2\cdot V=0,01\;\Rightarrow\;V=\SI{0,005}{\metre\cubed}=\SI{5}{\deci\metre\cubed} + \end{align*} + La température maximale est alors~: + \begin{align*} + &T(0,005)=-3\cdot 10^7 (0,005^2-0,01\cdot 0,005)\\ + &=\SI{750}{\kelvin}\\ + \end{align*} + \end{enumerate} \end{Solution OS} \begin{Solution OS}{38}