D.59 Analy­sis of the ideal gas Carnot cy­cle

Re­fer to fig­ure D.4 for the phys­i­cal de­vice to be an­a­lyzed. The re­frig­er­ant cir­cu­lat­ing through the de­vice is an ideal gas with con­stant spe­cific heats, like a thin gas of he­lium atoms. Chap­ter 11.14 will ex­am­ine ideal gases in de­tail, but for now some re­minders from in­tro­duc­tory clas­si­cal physics classes about ideal gasses must do. The in­ter­nal en­ergy of the gas is $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mIC_VT$ where $mI$ is its mass and $C_v$ is a con­stant for a gas like he­lium whose atoms only have trans­la­tional ki­netic en­ergy. Also, the ideal gas law says that $PV$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mIRT$, where $P$ is the pres­sure, $V$ the vol­ume, and the con­stant $R$ is the gas con­stant, equal to the uni­ver­sal gas con­stant di­vided by the mo­lar mass.

The dif­fer­en­tial ver­sion of the first law, en­ergy con­ser­va­tion, (11.11), says that

\begin{displaymath}
{\rm d}E= \delta Q - P{ \rm d}V
\end{displaymath}

or get­ting rid of in­ter­nal en­ergy and pres­sure us­ing the given ex­pres­sions,

\begin{displaymath}
mIC_v { \rm d}T = \delta Q - mIRT \frac{{\rm d}V}{V}.
\end{displaymath}

Fig­ure D.4: Schematic of the Carnot re­frig­er­a­tion cy­cle.
\begin{figure}\centering
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\begin{picture}(230,17...
...t(-72,48){\circle{14}}\put(-72,48){\makebox(0,0){4}}
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Now for the tran­si­tions through the heat ex­chang­ers, from 1 to 2 or from 3 to 4 in fig­ure D.4, the tem­per­a­ture is ap­prox­i­mated to be con­stant. The first law above can then be in­te­grated to give the heat added to the sub­stance as:

\begin{displaymath}
Q_{\rm {L}}= mIRT_{\rm {L}} \left(\ln V_2 - \ln V_1\right)
...
...d
Q_{\rm {H}}=-mIRT_{\rm {H}} \left(\ln V_4 - \ln V_3\right).
\end{displaymath}

Re­mem­ber that un­like $Q_{\rm {L}}$, $Q_{\rm {H}}$ is taken pos­i­tive if it comes out of the sub­stance.

On the other hand, for the tran­si­tions through the adi­a­batic tur­bine and com­pres­sor, the heat $\delta{Q}$ added is zero. Then the first law can be di­vided through by $T$ and in­te­grated to give

\begin{displaymath}
mI C_v \left(\ln T_{\rm {H}} - \ln T_{\rm {L}}\right)
= -mIR \left(\ln V_3 - \ln V_2\right)
\end{displaymath}


\begin{displaymath}
mI C_v \left(\ln T_{\rm {L}} - \ln T_{\rm {H}}\right)
= -mIR \left(\ln V_1 - \ln V_4\right)
\end{displaymath}

Adding these two ex­pres­sions shows that

\begin{displaymath}
\ln V_3 - \ln V_2 + \ln V_1 - \ln V_4 = 0
\quad \Longrightarrow\quad
\ln V_3 - \ln V_4 = \ln V_2 - \ln V_1
\end{displaymath}

and plug­ging that into the ex­pres­sions for the ex­changed heats shows that $Q_{\rm {H}}$$\raisebox{.5pt}{$/$}$$T_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q_{\rm {L}}$$\raisebox{.5pt}{$/$}$$T_{\rm {L}}$.