4.6 gm of liquid ethanol `(C_(2)H_(5)OH)` is taken in 12 litre container and at `27^(@)C`, 40% of ethanol is vaporised till equilibrium. Now if volume of container is halved and system is allowed to attain equilibrium then find [Give: R=0.08atm/litre mole-k]

Explanation

`C_(2)H_(5)OH(l)hArrC_(2)H_(5)OH(g)`
Molet=0 0.1 0
`t=t_("eqm") 0.1-0.04 0.04`
=0.06
For gas PV=nRT `rArrP_(C_(2)H_(5)OH("gas"))=(nRT)/(V)=(0.04xx0.08xx300)/(12)`
`V.P_("gas")=0.08` atm (at `eq^(m)`)
=0.08xx100=8