Respuesta :
Answer : The correct option is, (b) 20 kPa
Explanation :
The Raoult's law for liquid phase is:
[tex]p_A=x_A\times p^o_A[/tex] .............(1)
where,
[tex]p_A[/tex] = partial vapor pressure of A
[tex]p^o_A[/tex] = vapor pressure of pure substance A
[tex]x_A[/tex] = mole fraction of A
The Raoult's law for vapor phase is:
[tex]p_A=y_A\times p_T[/tex] .............(2)
where,
[tex]p_A[/tex] = partial vapor pressure of A
[tex]p_T[/tex] = total pressure of the mixture
[tex]y_A[/tex] = mole fraction of A
Now comparing equation 1 and 2, we get:
[tex]x_A\times p^o_A=y_A\times p_T[/tex]
[tex]p_T=\frac{x_A\times p^o_A}{y_A}[/tex] ............(3)
First we have to calculate the total pressure of the mixture.
Given:
[tex]x_A=0.4[/tex] and [tex]x_B=1-x_A=1-0.4=0.6[/tex]
[tex]y_A=0.7[/tex] and [tex]y_B=1-y_A=1-0.7=0.3[/tex]
[tex]p^o_A=70kPa[/tex]
Now put all the given values in equation 3, we get:
[tex]p_T=\frac{0.4\times 70kPa}{0.7}=40kPa[/tex]
Now we have to calculate the vapor pressure of B.
Formula used :
[tex]x_B\times p^o_B=y_B\times p_T[/tex]
[tex]p^o_B=\frac{y_B\times p_T}{x_B}[/tex]
Now put all the given values in this formula, we get:
[tex]p^o_B=\frac{0.3\times 40kPa}{0.6}=20kPa[/tex]
Therefore, the vapor pressure of B is 20 kPa.
