Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at `0^(@)C` at a pressure of `76cm` of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at `62^(@)C`.What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

Explanation

The temperature `(T_(1))` of gas when both flasks are immersed in boiling water is `100^(@)C` or `373 K` and pressure `(P_(1))` is `0.5atm`.
The average temperature of the gas when one flask is immersed in ice and the other in boiling water is
`T_(2)=(0+100)/(2)=50^(@)C=323 K`
As volume remains constant, according to Gay-Lussac's law,
`(P_(1))/(T_(1))=(P_(2))/(T_(2))`
`:. Pressure(P_(2))=(P_(1)xxT_(2))/(T_(1))=(0.5xx323)/(373)=0.433 atm`
Hence, pressure `P_(2)` is `0.433 atm`.