Question

Components A and B form ideal solution. At 350 K, a liquid mixture containing 40% (mole) A is in equilibrium with a vapour containing 70% (mole) A. If the vapour pressure of A at 350 K is 70 kPa, what is the vapour pressure of B? (b) 20 kPa (d) 12 kPa (а) 25 kPa (c) 40 kPa

Answer 1

Answer : The correct option is, (b) 20 kPa

Explanation :

The Raoult's law for liquid phase is:

`p_A=x_A* p^o_A` .............(1)

where,

`p_A` = partial vapor pressure of A

`p^o_A` = vapor pressure of pure substance A

`x_A` = mole fraction of A

The Raoult's law for vapor phase is:

`p_A=y_A* p_T` .............(2)

where,

`p_A` = partial vapor pressure of A

`p_T` = total pressure of the mixture

`y_A` = mole fraction of A

Now comparing equation 1 and 2, we get:

`x_A* p^o_A=y_A* p_T`

`p_T=(x_A* p^o_A)/(y_A)` ............(3)

First we have to calculate the total pressure of the mixture.

Given:

`x_A=0.4` and
`x_B=1-x_A=1-0.4=0.6`

`y_A=0.7` and
`y_B=1-y_A=1-0.7=0.3`

`p^o_A=70kPa`

Now put all the given values in equation 3, we get:

`p_T=(0.4* 70kPa)/(0.7)=40kPa`

Now we have to calculate the vapor pressure of B.

Formula used :

`x_B* p^o_B=y_B* p_T`

`p^o_B=(y_B* p_T)/(x_B)`

Now put all the given values in this formula, we get:

`p^o_B=(0.3* 40kPa)/(0.6)=20kPa`

Therefore, the vapor pressure of B is 20 kPa.