Fin de l'exercice avec la droite et le schéma pv
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@ -2903,7 +2903,38 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\item Trouvez la température maximale du gaz au cours de la compression de A à B.
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\end{enumerate}
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\begin{solos}
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Un autre corrigé de test.
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On commence par calculer le produit nR~:
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\[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\]
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\begin{enumerate}
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\item Puis, on calcule la chaleur échangée lors de la transformation~:
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\begin{align*}
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Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\
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&=5250+7350+2100=\SI{14700}{\joule}
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\end{align*}
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\item La transformation est linéaire. On peut la multiplier par le volume~:
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\begin{align*}
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p&=-3\cdot 10^8\cdot (V-0,01)\\
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pV&=-3\cdot 10^8\cdot (V-0,01)\cdot V
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\end{align*}
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Comme le gaz est parfait~:
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\begin{align*}
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&pV=-3\cdot 10^8\cdot (V-0,01)\cdot V=n\cdot R\cdot T\\
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&\Rightarrow\;T(V)=-3\cdot 10^7 (V^2-0,01\cdot V)
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\end{align*}
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car nR vaut dix.
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\item Pour trouver le maximum, il suffit alors de dériver et d'annuler~:
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\begin{align*}
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&\frac{dT}{dV}=-3\cdot 10^7 (2V-0,01)=0\\
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&\Rightarrow\;2\cdot V=0,01\;\Rightarrow\;V=\SI{0,005}{\metre\cubed}=\SI{5}{\deci\metre\cubed}
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\end{align*}
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La température maximale est alors~:
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\begin{align*}
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&T(0,005)=-3\cdot 10^7 (0,005^2-0,01\cdot 0,005)\\
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&=\SI{750}{\kelvin}\\
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\end{align*}
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\end{enumerate}
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\end{solos}
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\end{exos}
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@ -2914,6 +2945,13 @@ 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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% Un énoncé de test.
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% \begin{solos}
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% Un autre corrigé de test.
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% \end{solos}
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%\end{exos}
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}
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\subsection{Relatifs aux incertitudes}
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@ -2890,7 +2890,8 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\centering
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\caption{Le diagramme PV.\label{lediagpv}}
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\centering
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\input{Annexe-Exercices/Images/DiagPV.eps_tex}
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\def\svgwidth{0.9\columnwidth}
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\input{Annexe-Exercices/Images/DiagPV.ps_tex}
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\end{figure}
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\begin{enumerate}
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\item Calculez la chaleur reçue par le gaz pour passer de A à B.
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@ -2902,7 +2903,38 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\item Trouvez la température maximale du gaz au cours de la compression de A à B.
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\end{enumerate}
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\begin{solos}
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Un autre corrigé de test.
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On commence par calculer le produit nR~:
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\[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\]
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\begin{enumerate}
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\item Puis, on calcule la chaleur échangée lors de la transformation~:
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\begin{align*}
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Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\
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&=5250+7350+2100=\SI{14700}{\joule}
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\end{align*}
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\item La transformation est linéaire. On peut la multiplier par le volume~:
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\begin{align*}
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p&=-3\cdot 10^8\cdot (V-0,01)\\
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pV&=-3\cdot 10^8\cdot (V-0,01)\cdot V
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\end{align*}
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Comme le gaz est parfait~:
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\begin{align*}
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&pV=-3\cdot 10^8\cdot (V-0,01)\cdot V=n\cdot R\cdot T\\
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&\Rightarrow\;T(V)=-3\cdot 10^7 (V^2-0,01\cdot V)
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\end{align*}
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car nR vaut dix.
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\item Pour trouver le maximum, il suffit alors de dériver et d'annuler~:
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\begin{align*}
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&\frac{dT}{dV}=-3\cdot 10^7 (2V-0,01)=0\\
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&\Rightarrow\;2\cdot V=0,01\;\Rightarrow\;V=\SI{0,005}{\metre\cubed}=\SI{5}{\deci\metre\cubed}
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\end{align*}
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La température maximale est alors~:
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\begin{align*}
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&T(0,005)=-3\cdot 10^7 (0,005^2-0,01\cdot 0,005)\\
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&=\SI{750}{\kelvin}\\
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\end{align*}
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\end{enumerate}
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\end{solos}
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\end{exos}
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@ -2913,6 +2945,13 @@ 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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% Un énoncé de test.
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% \begin{solos}
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% Un autre corrigé de test.
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% \end{solos}
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%\end{exos}
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}
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\subsection{Relatifs aux incertitudes}
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@ -1,4 +1,4 @@
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@ -38,7 +38,7 @@
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70
Annexe-Exercices/Images/DiagPV.ps_tex.bak
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70
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<bcf:citekey order="8">SH03</bcf:citekey>
|
||||
<bcf:citekey order="9">SH03</bcf:citekey>
|
||||
<bcf:citekey order="10">EL99</bcf:citekey>
|
||||
<bcf:citekey order="11">EL99</bcf:citekey>
|
||||
<bcf:citekey order="12">GSJ05</bcf:citekey>
|
||||
<bcf:citekey order="13">SH03</bcf:citekey>
|
||||
<bcf:citekey order="14">BM85</bcf:citekey>
|
||||
<bcf:citekey order="15">BM85</bcf:citekey>
|
||||
<bcf:citekey order="16">BM85</bcf:citekey>
|
||||
<bcf:citekey order="17">SH03</bcf:citekey>
|
||||
<bcf:citekey order="18">OG04</bcf:citekey>
|
||||
<bcf:citekey order="19">BS07</bcf:citekey>
|
||||
<bcf:citekey order="20">GC88</bcf:citekey>
|
||||
<bcf:citekey order="21">GC88</bcf:citekey>
|
||||
<bcf:citekey order="22">LJ04</bcf:citekey>
|
||||
<bcf:citekey order="23">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="24">OP06</bcf:citekey>
|
||||
<bcf:citekey order="25">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="26">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="27">SV06</bcf:citekey>
|
||||
<bcf:citekey order="28">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="29">SU03</bcf:citekey>
|
||||
<bcf:citekey order="30">SU03</bcf:citekey>
|
||||
<bcf:citekey order="31">AK05</bcf:citekey>
|
||||
<bcf:citekey order="32">AS06</bcf:citekey>
|
||||
<bcf:citekey order="33">AS02</bcf:citekey>
|
||||
<bcf:citekey order="34">JJD21</bcf:citekey>
|
||||
<bcf:citekey order="35">JL96</bcf:citekey>
|
||||
<bcf:citekey order="36">GG92</bcf:citekey>
|
||||
<bcf:citekey order="37">JR07</bcf:citekey>
|
||||
<bcf:citekey order="38">GC88</bcf:citekey>
|
||||
<bcf:citekey order="39">JR00</bcf:citekey>
|
||||
<bcf:citekey order="40">GC88</bcf:citekey>
|
||||
<bcf:citekey order="41">GC88</bcf:citekey>
|
||||
<bcf:citekey order="42">BS07</bcf:citekey>
|
||||
<bcf:citekey order="43">HNHR08</bcf:citekey>
|
||||
<bcf:citekey order="44">JR00</bcf:citekey>
|
||||
<bcf:citekey order="45">FC05</bcf:citekey>
|
||||
<bcf:citekey order="46">SV06</bcf:citekey>
|
||||
<bcf:citekey order="47">FB05</bcf:citekey>
|
||||
<bcf:citekey order="48">JL06</bcf:citekey>
|
||||
<bcf:citekey order="49">SH03</bcf:citekey>
|
||||
<bcf:citekey order="50">EL99</bcf:citekey>
|
||||
<bcf:citekey order="51">GSJ05</bcf:citekey>
|
||||
<bcf:citekey order="52">BM85</bcf:citekey>
|
||||
<bcf:citekey order="53">OG04</bcf:citekey>
|
||||
<bcf:citekey order="54">BS07</bcf:citekey>
|
||||
<bcf:citekey order="55">GC88</bcf:citekey>
|
||||
<bcf:citekey order="56">LJ04</bcf:citekey>
|
||||
<bcf:citekey order="57">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="58">OP06</bcf:citekey>
|
||||
<bcf:citekey order="59">SU03</bcf:citekey>
|
||||
<bcf:citekey order="60">AK05</bcf:citekey>
|
||||
<bcf:citekey order="61">AS06</bcf:citekey>
|
||||
<bcf:citekey order="62">AS02</bcf:citekey>
|
||||
<bcf:citekey order="63">JJD21</bcf:citekey>
|
||||
<bcf:citekey order="64">JL96</bcf:citekey>
|
||||
<bcf:citekey order="65">GG92</bcf:citekey>
|
||||
<bcf:citekey order="66">JR07</bcf:citekey>
|
||||
<bcf:citekey order="1" intorder="1">HNHR08</bcf:citekey>
|
||||
<bcf:citekey order="2" intorder="1">JR00</bcf:citekey>
|
||||
<bcf:citekey order="3" intorder="1">FC05</bcf:citekey>
|
||||
<bcf:citekey order="4" intorder="1">SV06</bcf:citekey>
|
||||
<bcf:citekey order="5" intorder="1">FB05</bcf:citekey>
|
||||
<bcf:citekey order="6" intorder="1">SV06</bcf:citekey>
|
||||
<bcf:citekey order="7" intorder="1">JL06</bcf:citekey>
|
||||
<bcf:citekey order="8" intorder="1">SH03</bcf:citekey>
|
||||
<bcf:citekey order="9" intorder="1">SH03</bcf:citekey>
|
||||
<bcf:citekey order="10" intorder="1">EL99</bcf:citekey>
|
||||
<bcf:citekey order="11" intorder="1">EL99</bcf:citekey>
|
||||
<bcf:citekey order="12" intorder="1">GSJ05</bcf:citekey>
|
||||
<bcf:citekey order="13" intorder="1">SH03</bcf:citekey>
|
||||
<bcf:citekey order="14" intorder="1">BM85</bcf:citekey>
|
||||
<bcf:citekey order="15" intorder="1">BM85</bcf:citekey>
|
||||
<bcf:citekey order="16" intorder="1">BM85</bcf:citekey>
|
||||
<bcf:citekey order="17" intorder="1">SH03</bcf:citekey>
|
||||
<bcf:citekey order="18" intorder="1">OG04</bcf:citekey>
|
||||
<bcf:citekey order="19" intorder="1">BS07</bcf:citekey>
|
||||
<bcf:citekey order="20" intorder="1">GC88</bcf:citekey>
|
||||
<bcf:citekey order="21" intorder="1">GC88</bcf:citekey>
|
||||
<bcf:citekey order="22" intorder="1">LJ04</bcf:citekey>
|
||||
<bcf:citekey order="23" intorder="1">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="24" intorder="1">OP06</bcf:citekey>
|
||||
<bcf:citekey order="25" intorder="1">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="26" intorder="1">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="27" intorder="1">SV06</bcf:citekey>
|
||||
<bcf:citekey order="28" intorder="1">HRaFL03</bcf:citekey>
|
||||
<bcf:citekey order="29" intorder="1">SU03</bcf:citekey>
|
||||
<bcf:citekey order="30" intorder="1">SU03</bcf:citekey>
|
||||
<bcf:citekey order="31" intorder="1">AK05</bcf:citekey>
|
||||
<bcf:citekey order="32" intorder="1">AS06</bcf:citekey>
|
||||
<bcf:citekey order="33" intorder="1">AS02</bcf:citekey>
|
||||
<bcf:citekey order="34" intorder="1">JJD21</bcf:citekey>
|
||||
<bcf:citekey order="35" intorder="1">JL96</bcf:citekey>
|
||||
<bcf:citekey order="36" intorder="1">GG92</bcf:citekey>
|
||||
<bcf:citekey order="37" intorder="1">JR07</bcf:citekey>
|
||||
<bcf:citekey order="38" intorder="1">GC88</bcf:citekey>
|
||||
<bcf:citekey order="39" intorder="1">JR00</bcf:citekey>
|
||||
<bcf:citekey order="40" intorder="1">GC88</bcf:citekey>
|
||||
<bcf:citekey order="41" intorder="1">GC88</bcf:citekey>
|
||||
<bcf:citekey order="42" intorder="1">BS07</bcf:citekey>
|
||||
</bcf:section>
|
||||
<!-- SORTING TEMPLATES -->
|
||||
<bcf:sortingtemplate name="nty">
|
||||
|
||||
Binary file not shown.
@ -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}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user