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.
Solution
Let the initial pressure of the chambers A and B be PA1 and PB1 , respectively .
Let the final pressure of chambers A and B be PA2 and PB2 , respectively .
Let the CSA be A.
VA1 = 0.20 A
TA1 = 400 K
VB1 = 0.1 A
TB1 = 100 K
At first equilibrium , both side pressures will be the same .
⇒ PA1 = PB1
Let the final temperature at equilibrium be T .then ,
`(P_(A1)V_(A1))/(T_(A1)) =( P_(A2)V_(A2))/T`
⇒ `(P_(A1)0.2A)/400` = `(P_(A2)V_(A2))/T`
⇒ `P_(A2 )=( P_(A1)0.2AT)/(400V_(A2))` ......(1)
For Second chamber :
`(P_(B1)V_(B1))/(T_(B1))` = `(P_(B2)V_(B2))/T`
⇒ `(P_(B1)0.1A)/100 = (P_(B2)V_(B2))/T`
⇒ `P_(B2) = (P_(B1)0.1AT)/(100V_(B2))`....(2)
At Second equilibrium , pressures on both sides will be the same .
⇒ `P_(A2) = P_(B2)`
⇒ `(P_(A1)0.2AT)/(400V_(A2)) = (P_(B1)0.1AT)/(100V_(B2))`
⇒ `P_(A1)/(2V_(A2)) = P_(B1)/V_(B2)`
⇒ `V_(B2) = 2V_(A2)` ...........(3)
Now ,
`V_(B2) + V_(A2) = 0.3A`
⇒ `3V_(A2) = 0.3A`
⇒ `V_(A2) = 0.1A`
Let `V_(A2)` be lA.
⇒ l = 0.1 m
⇒ l = 10 cm
