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Two monconducting containers having volu...

Two monconducting containers having volume `V_(1)` and `V_(2)` contain monoatomic and dimatomic gases respectivley. They are connected as shown in figure. Pressure and temperature in the two containers are `P_(1), T_(1)` and `P_(2), T_(2)` respectively. Initially stop cock is closed, if the stop coc is opened find the final pressure and temperature.

Text Solution

Verified by Experts

`n_(1) = (P_(1)V_(1))/(RT_(1)) n_(2) = (P_(2)V_(2))/(RT_(2))`
`n = n_(1) +n_(2)` (number of moles are conserved)
Finally pressure in both parts & temperature of the both the gases will becomes equal.
`(P(V_(1)+V_(2)))/(RT) = (P_(1)V_(1))/(RT_(1)) +(P_(2)V_(2))/(RT_(2))`
From energy conservation
`(3)/(2)n_(1)RT_(1)+(5)/(2) n_(2)RT_(2) = (3)/(2)n_(1)RT + (5)/(2) n_(2)RT`
`rArr T = ((3P_(1)V_(1)+5P_(2)V_(2))T_(1)T_(2))/(3P_(1)V_(1)T_(2)+5P_(1)V_(2)T_(1)) rArrP =((3P_(1)V_(1)+5P_(2)V_(2))/(3P_(1)V_(1)T_(2)+5P_(1)V_(2)T_(1)))((P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))/(V_(1)+V_(2)))`
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Knowledge Check

  • Two adiabatic containers have volumes V_(1) and V_(2) respectively. The first container has monoatomic gas at pressure p_(1) and temperature T_(1) . The second container has another monoatomic gas at pressure p_(2) and temperature T_(2) . When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. Then

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