Question

Let ( P ) be the total pressure. Let ( X ) be the number of moles of ( boldsymbol{A} ) in liquid state.
The number of moles in vapour state will be
( 10-x )
Let ( boldsymbol{y} ) be the number of moles of ( boldsymbol{B} ) in liquid state
The number of moles of ( boldsymbol{B} ) in vapour state will be ( 10-y )
Hence, ( x+y=10 )
( P times frac{(10-x)}{(10-x+10-y)}=200 times frac{(x)}{(x+y)} )
( P times frac{(10-y)}{(10-x+10-y)}=frac{100 times(y)}{(x+y)} )
These equations are solved for ( x, y ) and ( P ).
( x=4.142 )
( y=5.858 )
( P=141.4 mathrm{mm} mathrm{Hg} )

# Pressure over ideal binaryliquid mixture containing 10 moles each of liquid Aand Bis gradually decrease isothermally. If P&=200 mm Hg and Po=100 mm Hg, find the pressure at which half of the liquid is converted into vapour. (A) 150 mm Hg (B) 166.5 mm Hg (C) 133 mm Hg (D) 141.4 mm Hg

Solution