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NEET 2012
Exam
NEET
Year
2012
Subject
Chemistry
Chapter
Solutions

NEET 2012 · Chemistry · Solutions

p_A and p_B are the vapour pressure of pure liquid components, A...

pAp_A and pBp_B are the vapour pressure of pure liquid components, AA and BB, respectively of an ideal binary solution. If xAx_A represents the mole fraction of component AA, the total pressure of the solution will be

Options

  1. ApA+xA(pB−pA)p_A+x_A\left(p_B-p_A\right)
  2. BpA+xA(pA−pB)p_A+x_A\left(p_A-p_B\right)
  3. CpB+xA(pB−pA)p_B+x_A\left(p_B-p_A\right)
  4. DpB+xA(pA−pB)p_B+x_A\left(p_A-p_B\right)

Correct answer

D. pB+xA(pA−pB)p_B+x_A\left(p_A-p_B\right)

Step-by-step solution

Total pressure, pT=pA′+pB′p_T=p_A^{\prime}+p_B^{\prime} We know that, pA′=pAxApB′=pBxBp_A^{\prime}=p_A x_A p_B^{\prime}=p_B x_B Substituting the values of pA′andpB′p_A^{\prime} and p_B^{\prime} in Eq. (i) pT=pAxA+pBxB[xA+xB=1⇒xA=1−xB or xB=1−xA]=pAxA+pB(1−xA)=pAxA+pB−pBxA∴pT=pB+xA(pA−pB)\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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