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$p_A$ and $p_B$ are the vapour pressure of pure liquid components, $A$ and $B$, respectively of an ideal binary solution. If $x_A$ represents the mole fraction of component $A$, the total pressure of the solution will be
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Verified Answer
The correct answer is:
$p_B+x_A\left(p_A-p_B\right)$
Total pressure,
$p_T=p_A^{\prime}+p_B^{\prime}$
We know that, $p_A^{\prime}=p_A x_A$
$p_B^{\prime}=p_B x_B$
Substituting the values of $p_A^{\prime}$ and $p_B^{\prime}$ in Eq. (i)
$\begin{aligned}
p_T & =p_A x_A+p_B x_B \\
{\left[x_A+x_B\right.} & \left.=1 \Rightarrow x_A=1-x_B \text { or } x_B=1-x_A\right] \\
& =p_A x_A+p_B\left(1-x_A\right) \\
& =p_A x_A+p_B-p_B x_A \\
\therefore p_T & =p_B+x_A\left(p_A-p_B\right)
\end{aligned}$
$p_T=p_A^{\prime}+p_B^{\prime}$
We know that, $p_A^{\prime}=p_A x_A$
$p_B^{\prime}=p_B x_B$
Substituting the values of $p_A^{\prime}$ and $p_B^{\prime}$ in Eq. (i)
$\begin{aligned}
p_T & =p_A x_A+p_B x_B \\
{\left[x_A+x_B\right.} & \left.=1 \Rightarrow x_A=1-x_B \text { or } x_B=1-x_A\right] \\
& =p_A x_A+p_B\left(1-x_A\right) \\
& =p_A x_A+p_B-p_B x_A \\
\therefore p_T & =p_B+x_A\left(p_A-p_B\right)
\end{aligned}$
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