0
$\begingroup$

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
-1
$\begingroup$

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 $$2nC_vdT=nRTd\ln{\left[\frac{(1+\eta)^2}{\eta}\right]}$$

$\endgroup$

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.