(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 ;)


2 Answers 2


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.

  • $\begingroup$ What happened to the n that is supposed to be in front of the Cv? $\endgroup$ Aug 20, 2019 at 13:28
  • $\begingroup$ as in why am I starting with $C_vdT$ instead of $nC_vdT$? $\endgroup$ Aug 20, 2019 at 13:38
  • $\begingroup$ Yes, that's what I'm asking. $\endgroup$ Aug 20, 2019 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 $\endgroup$ Aug 20, 2019 at 13:43
  • 1
    $\begingroup$ @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, 2019 at 6:40

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 $$\frac{T_x}{T_y}=\left(\frac{P_x}{P_y}\right)^{(\gamma-1)/\gamma}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.