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

$${\rm d}E= \delta Q - P{ \rm d}V$$

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

$$mIC_v { \rm d}T = \delta Q - mIRT \frac{{\rm d}V}{V}.$$

Fig­ure D.4: Schematic of the Carnot re­frig­er­a­tion cy­cle.

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:

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

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

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

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

Adding these two ex­pres­sions shows that

$$\ln V_3 - \ln V_2 + \ln V_1 - \ln V_4 = 0 \quad \Longrightarrow\quad \ln V_3 - \ln V_4 = \ln V_2 - \ln V_1$$

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}}$.