2
$\begingroup$

A heat conducting piston can freely move inside a closed thermally insulated cylinder with an ideal gas. In equilibrium the piston divides into two equal parts, the gas temperature being equal to $T_o$ . The piston is slowly displaced. Find the gas temperature as a function of the ratio $\eta$ of the greater and smaller sections. The adiabatic exponent of the gas is equal to $\gamma$.

I am tired of trying this problem over and over again. Here's my attempt:

At any instant, let $V_1$ be the volume of the smaller section and $V_2$ be the volume of the larger section. The presures are $p_1$ and $p_2$ respectively.l

The temperature on both sides of the piston is same and equal to $T$ because the diathermal piston moves slowly.

$p_1V_1 = p_2 V_2 \implies p_1 V_1= p_2 \eta V_1 \implies p_1 = \eta p_2 \implies \\ dW = (p_2-p_1)dV = (p_2(1-\eta))dV $

Also,

$\dfrac{V_2}{V_1}= \eta$

Let $C_v$ be the molar heat capacity at constant volume. Also, let there be $n$ moles each side.

Now, for the entire system $dU = -dW$ from the first law of thermodynamics.

$$2nC_v dT = p_2(\eta -1)dV$$

Note that $$p_2 =\dfrac{ nrT_2}{V_2} $$

$$\implies \dfrac{2C_v}{R}\dfrac{1}{T}dT= (\dfrac{\eta}{V_2}- \dfrac{1}{V_2})dV$$

Now, clearly $\eta/ V_2 = 1/V_1$

$$\implies \int ^T_{T_o} \dfrac{2C_v}{R}\dfrac{1}{T}dT= \left(\int^{V_1}_{V_o}\dfrac{1}{V_1}dV- \int ^{V_2} _{V_0}\dfrac{1}{V_2}dV\right)$$

On solving,

$$\implies T= T_o\left[\dfrac{1}{\eta}\right]^{\dfrac{\gamma- 1}{2}}$$ Because $C_v = \dfrac{R}{\gamma -1 }$

However, the answer is: $$T= T_o\left[\dfrac{(\eta+1)^2}{4\eta}\right]^{\dfrac{\gamma-1}{2}}$$

Please let me know the conceptual error I have made. Thank you.

$\endgroup$
  • 1
    $\begingroup$ By $dV$, you appear to mean $dV_2$, so the integral $\int dV_2/V_1$ requires you to express $V_1$ in terms of $V_2$. That should help, I think - up to that point everything seems fine. $\endgroup$ – J. Murray Feb 14 '18 at 17:44
  • $\begingroup$ @J.Murray isn't $dV_1 = dV_2$? $\endgroup$ – Archer Feb 14 '18 at 18:03
  • $\begingroup$ No. You've already (correctly) used the fact that $dV_1=-dV_2$ in your first line. $\endgroup$ – J. Murray Feb 14 '18 at 18:06
  • $\begingroup$ How can you say that the energy change of whole system is equal to work done by one single gas? don't you need the net work $\endgroup$ – Buraian Sep 8 at 11:18
  • $\begingroup$ and, secondly, how did you take $ T_2 to the other side? $\endgroup$ – Buraian Sep 12 at 11:05
0
$\begingroup$

I get $$dW=nRTd\ln{V_1}+nRTd\ln{V_2}=nRTd\ln{(V_1V_2)}$$ where $$V_1=\frac{2V_0}{(1+\eta)}$$and $$V_2=\frac{2V_0\eta}{(1+\eta)}$$This leads directly to $$2nC_vdT=nRTd\ln{\left[\frac{(1+\eta)^2}{\eta}\right]}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hello Mr. Miller, I tried doing this question by using the adiabatic law at the initial equilibrium state and the final equilibrium state. $$T_1 (V_1)^{\gamma -1} = T_2 (V_2)^{\gamma -1}$$ and tried to figure out volumes from that. I took inital volume as $ \frac{V_{net}}{2}$ and final volume as $ \frac{2}{3} V_{net}$ $\endgroup$ – Buraian Sep 8 at 11:00
  • $\begingroup$ I'd be glad if you could point out the error which exists in my attempt $\endgroup$ – Buraian Sep 8 at 11:01
  • $\begingroup$ Your error is in evaluating the first integral on the right hand side of the equation (involving V1). $\endgroup$ – Chet Miller Sep 8 at 21:10

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