Two bulbs of 100 c.c and 200 c.c capacity, contain the same gas at same temperature and pressure are connected by a capillary tube of negligible volume. Initially both were at `0^(@)C`. Now the temperature of the bigger bulb is raised to `100^(@)C` and that of the smaller bulb being kept at `0^(@)C`. If the initial pressure was 76 cm of mercury, what is the final pressure ?

Explanation

Let P be the initial pressure of the bulbs and `P^(1)` be the final pressure. Initial temperature of the bulbs be T. The final temperatures of the bulbs of volume `V_(1)andV_(2)" be "T_(1)andT_(2)` respectively. By ideal gas equation, as mass of the gas in the bulbs remains constant `(P_(1)V_(1))/T+(P_(2)V_(2))/T=(P^(1)V_(1))/T_(1)+(P^(1)V_(2))/T_(2)`
`P/T(V_(1)+V_(2))=P^(1)(V_(1)/T_(1)+V_(2)/T_(2))" "(becauseP_(1)=P_(2)=P)`
`76/273(100+200)=P^(1)(100/273+200/373)`
`(76xx300)/273=P^(1)xx100(99/(273xx373))`
`P^(1)=(76xx3xx373)/919=92.53cm`