Two glass bulbs of volumes 3L and 1L are connected by a narrow tube. The system is filled with air at`30^@C` temperature and at 76 cmHg pressure. Now the bulb of volume 3L is immersed in water vapour at temperature `100^@C` while the other bulb is kept at `30^@C`. Find the air pressures in the two bulbs.Neglect the volume expansion of the 3L bulb.

Explanation

Initially the total volume of the bulbs (V)=3+1=4L air pressure `(p_1)=76 cmHg` temperature of air `(T_1)=273+30=303K`
In the 2nd case air pressure =`p_2`
Temperature of 3L bulb =`100+273=373K`
temperature of 1L bulb=30+273=303 K
We use the relation, `n=(pV)/(RT)`
In the 1st case number of gram-molecules in the gas
`=(76times3)/(R times 303)+(76times1)/(R times 303)=76/(R times 303)(3+1)=(76times4)/(R times 303)`
let in the 2nd case air pressure=p
No. of gram molecules `=(ptimes3)/(R times373)+(ptimes1)/(R times303)=p/R(3/373+1/303)`
`because` No.of gram molecules is unchanged
`(76times4)/(R times303)=p/R(3/373+1/303)or,(76times4)/303=p((909+373)/(373times303))`
or, `p=(76times4times373)/1282=88.4cmHg.`