1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you tell me with which equation you are not familiar and where you found the typos? $\endgroup$ – Andreas Oct 15 '13 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) $\endgroup$ – Krishna Oct 15 '13 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 '13 at 10:31
  • $\begingroup$ I edited my answer. Your assumption at the top were off by a - sign, it seems. $\endgroup$ – Krishna Oct 15 '13 at 11:01
  • 1
    $\begingroup$ $$C_vd T = - pdV$$ $\endgroup$ – mcodesmart Oct 15 '13 at 15:45
0
$\begingroup$

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 conventian, 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 $C_p/C_v$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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