Question

Figure shows a horizontal cylindical container of length 30cm, which is partitioned by a tight fitting separator. The separator is diathermic but conducts hear very slowly Initially the separator is in the state shown in figure. The temperature of left part of ylinder is 100K and that on right part is 400k. Initially the separator is in equilibrium. As heat is conducted from right to left part, seprator displaces to the right Find the displacment of separator after a long time, when gases on the parts of cylinder are in thermal equilibrium.​

Answer 1

AnSwer :-

It is given that initially the separator is in equilibrium thus Pressures of the gas on both sides of the separator are equal say, it is P$\sf _i$. If A be the area of cross - section of cylinder, number of moles of gas in left and right parts, n₁ and n₂ can be given as

$\sf n_1 = \dfrac{P_i (10A)}{R(100)} \: \: \: \: and \: \: \: \: n_2 = \dfrac{P_i (20A)}{R(400)}$

Finally, if separator is displaced to right by a distance x, we have

$\sf n_1 = \dfrac{P_f (10 + x)A}{RT_f} \: \: and \: \: n_2 = \dfrac{P_f (20 + x)A}{RT_f}$

Where P$\sf _f$ and T$\sf _f$ be the final Pressure and temperature on both sides after a long time.

Now if we equate the ratio of moles $\sf \dfrac{n_1}{n_2}$ in initial and final States, we get

$\sf\dfrac{n_1}{n_2} = \dfrac{(10A/100)}{(20A/400)} = \dfrac{(10 + x)A}{(20 - x)A}$

$\sf 2(20 - x) = 10 + x$

$\sf 40 - 2x = 10 + x$

$\sf 40 - 10 = x + 2x$

$\sf 3x = 30$

$\sf x = 10cm$

thus in final state, when gases in both parts are in thermal equilibrium, the pistion is displaced to 10cm right from its initial position.

$\:$

#sanvi....

Answer 2

It is given that initially the separator is in equilibrium thus Pressures of the gas on both sides of the separator are equal say, it is P\sf _ii . If A be the area of cross - section of cylinder, number of moles of gas in left and right parts, n₁ and n₂ can be given as

\sf n_1 = \dfrac{P_i (10A)}{R(100)} \: \: \: \: and \: \: \: \: n_2 = \dfrac{P_i (20A)}{R(400)}n1=R(100)Pi(10A)andn2=R(400)Pi(20A)

Finally, if separator is displaced to right by a distance x, we have

\sf n_1 = \dfrac{P_f (10 + x)A}{RT_f} \: \: and \: \: n_2 = \dfrac{P_f (20 + x)A}{RT_f}n1=RTfPf(10+x)Aandn2=RTfPf(20+x)A

Where P\sf _ff and T\sf _ff be the final Pressure and temperature on both sides after a long time.

Now if we equate the ratio of moles \sf \dfrac{n_1}{n_2}n2n1 in initial and final States, we get

\sf\dfrac{n_1}{n_2} = \dfrac{(10A/100)}{(20A/400)} = \dfrac{(10 + x)A}{(20 - x)A}n2n1=(20A/400)(10A/100)=(20−x)A(10+x)A

\sf 2(20 - x) = 10 + x2(20−x)=10+x

\sf 40 - 2x = 10 + x40−2x=10+x

\sf 40 - 10 = x + 2x40−10=x+2x

\sf 3x = 303x=30

\sf x = 10cmx=10cm

thus in final state, when gases in both parts are in thermal equilibrium, the pistion is displaced to 10cm right from its initial position.