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(for isentropic, adiabatic, ideal gas flow) $$\frac{T_x}{T_y} = \left(\frac{p_x}{p_y}\right)^{\frac{(\gamma-1)}{\gamma}} = \left(\frac{V_y}{V_x}\right)^{\gamma-1}$$ where $V$ is the specific volume, $x$ and $y$ are any 2 rocket nozzle axial sections (though I presume this relation holds true for many other things as well), and $\gamma$ is the specific heat capacity ratio $c_p/c_v$, numerically equivalent to the molar heat capacity ratio $(\bar{C}_p/\bar{C}_v)$.

Thanks! p.s. I may have just learnt the hard way that it is probably not the best idea to write a markdown-dense blurb on a phone ;)

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For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:

$$\mathrm dU = \delta Q + \delta W$$

$$\mathrm dU = - p\,\mathrm dV$$

This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...

$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$

$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$

$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$

$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$

From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.

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  • $\begingroup$ What happened to the n that is supposed to be in front of the Cv? $\endgroup$ – Chet Miller Aug 20 at 13:28
  • $\begingroup$ as in why am I starting with $C_vdT$ instead of $nC_vdT$? $\endgroup$ – Michael Green Aug 20 at 13:38
  • $\begingroup$ Yes, that's what I'm asking. $\endgroup$ – Chet Miller Aug 20 at 13:42
  • $\begingroup$ Because the constant volume heat capacity is defined as $C_v = (\frac{\partial U}{\partial T})_V$, whereas $C_{v_m} = \frac{1}{n} (\frac{\partial U}{\partial T})_V$ when utilizing the molar volume $\endgroup$ – Michael Green Aug 20 at 13:43
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    $\begingroup$ @ChetMiller: Among the physical chemistry textbooks I've used, $C_V$ refers to the total constant-V heat capacity of the substance or system (this is also the IUPAC convention: goldbook.iupac.org/terms/view/H02753 ) and a tilde, subscript-m, or some other designation (e.g., $\tilde{C_V}$, $C_{V,m}$) is used to indicate that the value is a molar constant-V heat capacity. $\endgroup$ – theorist Aug 21 at 6:40
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From the open system (control volume) version of the first law of thermodynamics, between cross sections x and y, $$\dot{Q}-\dot{W}_s-\dot{m}\Delta h=0$$where $\dot{Q}$ is the rate of heat addition to the control volume, $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}$ is the mass flow rate, and $\Delta h$ is the change in specific enthalpy between cross sections x and y. If the process is adiabatic and reversible, no shaft work is being done, and the gas is ideal, this equation reduces to $$dh=C_pdT=VdP=\frac{RT}{P}dP$$ along the path between the two cross sections, where V is the molar volume. This equation integrates to $$\frac{T_x}{T_y}=\left(\frac{P_x}{P_y}\right)^{(\gamma-1)/\gamma}$$

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