Why is there a sign error in my derivation? Adiabatic expansion of ideal gas: $TV^{1-\gamma}$ is constant 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.
 A: 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.
Edit
Ok, $dU + pdV = 0 \Rightarrow C_{v}d\tau + pdV = 0$ So, the culprit is at the top.
A: 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 $C_p/C_v$
