Question

Figure ( $24 - E 4$ ) 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 temperature in 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.

Accepted Answer

Let the area of both sides be $A$ ,

$P_{A}$ and $P_{B}$ are pressures on $400 K$ and $B$ .

The initial temperature,

$(T_{A}) = 400 K$

The initial temperature,

$(T_{B}) = 100 K$

At the equilibrium position, the force on $A$ is equal to force on $B$ ,

$\Rightarrow F_{A} = F_{B}$
$\Rightarrow P_{A} (A) = P_{B} (A)$
$\Rightarrow P_{A} = P_{B}$

As the separator is weakly conducting, the heat exchange will take place until the temperature $(T)$ in the final condition is equal, after a long time.

Let the final pressure $P$ be the equilibrium pressure.

Let the final position of the separating wall be at a distance $x$ from the left hand and $(30 - x)$ from the right.

Using the ideal gas equation for side $A$ ,

$P_{A} V_{A} = n_{A} R T_{A}$ ……. (i)

For $B$ ,

$400 K$ ……. (ii)

As we know, the number of moles,

$n_{A} = n_{B}$

Comparing the initial and the final state of $A$ ,

$\frac{P_{A} V_{A}}{T_{A}} = \frac{P V_{A}'}{T}$

Where $V_{A}^{'}$ is the final volume in $A$ ,

$= (30 - x) A$

$V_{A} = 20 \times A$ is the initial volume of $A$ .

$\Rightarrow \frac{P_{A} (A \times 20)}{400} = \frac{P A (30 - x)}{T}$

$\Rightarrow \frac{P_{A}}{P} = \frac{20 (30 - x)}{T}$ ……. (i)

Similarly, for part $B$ ,

$\frac{P_{B} V_{B}}{T_{B}} = \frac{P {V_{B}}^{'}}{T}$

Where $V_{B}^{'}$ is the final volume of $B$ ,

$= A x$

The initial volume of $B$ ,

$= 10 A$

$\Rightarrow \frac{P_{B} \times 10 \times A}{100} = \frac{P \times x \times A}{T}$

$\Rightarrow \frac{P_{B}}{P} = \frac{10 x}{T}$ ……. (ii)
Now, finding the ratio of $\frac{(i)}{(i i)}$ ,

$\frac{P_{A} / P}{P_{B} / P} = 1 = \frac{20 (30 - x)}{10 x} (∵ P_{A} = P_{B})$

$\Rightarrow 10 x = 600 - 20 x$

$\Rightarrow 30 x = 600$

$x = \frac{600}{30} = 20$
Hence, the separator will be at a distance of $30 - x = 30 - 20 = 10 cm$ from the left end.

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