Question

Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.
Figure

Answer 1

The final equilibrium position of the sector I = 10 cm

Explanation:

Let the chambers A and B starting pressure be $P_A_1$ and $P_B_1$ respectively.  

Let chambers A and B end pressure be $P_A_2$ and $P_B_2$ respectively.

Let the cross sectional area be A.  

$\begin{aligned}&\mathrm{V}_{\mathrm{A} 1}=0.20 \mathrm{A}\\&T_{A 1}=400 K\end{aligned}$

$\begin{aligned}&\mathrm{V}_{\mathrm{B} 1}=0.1 \mathrm{A}\\&T_{51}=100 K\end{aligned}$

Both side-pressures will be the same at first equilibrium.

$P_A_1$ = $P_B_1$

Let the final equilibrium temperature be T.

Then,

$\frac{P_{A 1} V_{A 1}}{T_{A 1}}=\frac{P_{A 2} V_{A 2}}{T}$

$\frac{P_{A 1} 0.2 A}{400}=\frac{P_{A 2} V_{A 2}}{T}$

$P_{A 2}=\frac{P_{A 1} 0.2 A T}{400 V_{A 2}}$ ……….(1)

Chamber 2:

$\begin{aligned}&\frac{P_{B 1} V_{B 1}}{T_{B 1}}=\frac{P_{B 2} V_{B N}}{T}\\&\frac{P_{B 1} 0.1 A}{100}=\frac{P_{B 2} V_{B 2}}{T}\end{aligned}$

$P_{B 2}=\frac{P_{B 1} 0.1 A T}{100 V_{B 2}}$         ……….(2)

Pressures on both sides will be the same at second equilibrium.

$P_{A 2}=P_{B 2}$

$\frac{P_{A 1} 0.2 A T}{400 V_{A 2}}=\frac{P_{B 1} 0.1 A T}{100 V_{B 2}}$

$\frac{P_{A 1}}{2 A T}=\frac{P_{B 1}}{V_{B 2}}$             ..…… (3)

Now,

$V_{B 2}+V_{A 2}=0.3 A$

$3 \mathrm{V}_{\mathrm{A} 2}=0.3 \mathrm{A}$

$\mathrm{V}_{\mathrm{A} 2}=0.1 \mathrm{A}$

Let $V_A_2$ be I

l = 0.1 m

l = 10 cm

Therefore the final equilibrium position of the separator is 10 cm, reached after a long time $\text {Let V_{A 2} be lA_{2}}$