This is the last part of a derivation of the equation for an ideal gas undergoing reversible adiabatic expansion. I'm trying to prove that $T V^{\gamma-1}$ is constant, but my result is that $T V^{1-\gamma}$ is constant. Where am I going wrong?

I don't know what notation is standard, so please tell me if I should clarify something. I think this is just a mathematical mistake though.

$$C_vd T = pdV = \frac{NT}{V}dV$$ $$\frac{dT}{T} = \frac{N}{C_V}\frac{dV}{V} = \frac{C_p - C_V}{C_V}\frac{dV}{V}$$ $$\frac{dT}{T} = (\gamma-1) \frac{dV}{V}$$ $$d\ln T = (\gamma-1) d \ln V $$

Integrate from $T_0$ to $T_1$ and $V_0$ to $V_1$ (start state to end state of the process):

$$\ln T_1 - \ln T_0 = (\gamma-1)( \ln V_1 - \ln V_0 ) $$ $$\ln \frac{ T_1 }{T_0} = (\gamma-1)\ln \frac{ V_1}{ V_0 } $$ $$\frac{ T_1 }{T_0} = \left(\frac{ V_1}{ V_0 }\right)^{\gamma-1} $$

Now put the start state and the end state on different sides:

$$ \frac{T_1}{V_1^{\gamma-1}} = \frac{T_0}{V_0^{\gamma-1}} $$ $$ TV_1^{1-\gamma} = T_0 V_0^{1-\gamma} $$

Therefore, $T V^{1-\gamma}$ must be constant throughout the process.


2 Answers 2


I'm not quite familiar with the equation that you have here. There are couple of typos too. What I remember from these problems is that you first prove $PV^{\gamma}=constant$.

$PV=nRT \Rightarrow P=\frac{nRT}{V}$

So, $\frac{nRT}{V}V^{\gamma}=const. \Rightarrow T V^{\gamma-1}=const. $ since nR is constant for a given amount of gas.


Ok, $dU + pdV = 0 \Rightarrow C_{v}d\tau + pdV = 0$ So, the culprit is at the top.

  • $\begingroup$ Could you tell me with which equation you are not familiar and where you found the typos? $\endgroup$
    – Andreas
    Oct 15, 2013 at 10:15
  • $\begingroup$ edited my comment. One of the typo I was referring to is d in front of ln(\tau) and ln(V), which should just be ln(\tau) and ln(V) $\endgroup$
    – timmur
    Oct 15, 2013 at 10:19
  • $\begingroup$ I intended the d to be there since $\frac{1}{y}\frac{dy}{dx} = \frac{d\ln{y}}{dx}$ (if you're not too exact about the notation). What do you mean by "it should be $T_1V_0^{1-\gamma}$"? How did you get there? $\endgroup$
    – Andreas
    Oct 15, 2013 at 10:31
  • $\begingroup$ I edited my answer. Your assumption at the top were off by a - sign, it seems. $\endgroup$
    – timmur
    Oct 15, 2013 at 11:01
  • 1
    $\begingroup$ $$C_vd T = - pdV$$ $\endgroup$
    – mcodesmart
    Oct 15, 2013 at 15:45

I think the only problem is the initial rewriting of the first principle of Thermodynamics; actually it's only a matter of conventions, but remaining consistent with the internal convention, there result must not change.

Anyway, from the First principle of Thermodynamics, we have

\begin{equation} dU=\delta Q-\delta W \end{equation}

but we already know that $\delta Q=0$ since we are dealing with an adiabatic process. We then have $$ dU+\delta W =0\Rightarrow nC_vd\theta + pdV=0. $$

From the perfect gas equation of state, we can then rewrite $p$ in function of $V$, i.e., $p=\frac{nR\theta}{V}$.

On the other side, combining the two equations, we are led to the following differential equations:

$$\frac{nC_v}{\theta}d\theta = -\frac{nR\theta}{V}dV .$$

If we confine to the case of a quasi-static, reversible transformation we are allowed to integrate both sides $$ \int_{\theta_i}^{\theta}\frac{C_v}{\tau}d\tau = -\int_{V_i}^{V}\frac{R}{\nu}d\nu \\ \log\left(\frac{\theta}{\theta_i}\right)C_v = \log\left(-\frac{V}{V_i}\right)R $$

And so using only algebraic manipulations we can easily get $$\theta V^{R/C_v}=\theta_i V_i^{R/C_v}=cK^{R/C_v}=const $$ Now recalling that $R = C_p-C_v$ we can easily recover the factor $\gamma-1 = \frac{C_p-C_v}{C_v}$ putting $\gamma$ to be $C_p/C_v$


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