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Standard XII

Physics

Time Period of Pendulum

The vapour pr...

Question

The vapour pressure of the solution of two liquids $A (p^{o} = 80 m m$ ) and $B (p^{o} = 120 m m)$ is found to be $100 m m$ when $x_{A} = 0.4$ . The result shows that:

solution exhibits ideal behaviour

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solution shows positive deviations

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solution shows negative deviations

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solution will show positive deviations for lower concentration and negative deviations for higher concentrations

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Solution

The correct option is C solution shows negative deviations
$\begin{matrix} P_{A}^{0} & = 80 m m P_{B}^{\circ} = 120 m m x_{n} = 0.4 x_{B} = 0.6 P_{T} & = x_{A} P_{A}^{0} + x_{B} P_{0}^{0} = 0.4 \times 80 + 0.6 \times 120 = 32 + 72 P_{T} & = 104 \end{matrix}$
Since the given total pressure is less than the calculated Total pressure therefore, it shows a negative deviation hence option C is correct.

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Q. CASE1: Read the passage given below and answer the following question.

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The vapour pressure of the solution of two liquids A(po=80mm) and B(po=120mm) is found to be 100mm when xA=0.4. The result shows that:

The correct option is C solution shows negative deviationsP0A=80 mmP∘B=120 mmxn=0.4xB=0.6PT=xAP0A+xBP00=0.4×80+0.6×120=32+72PT=104 Since the given total pressure is less than the calculated Total pressure therefore, it shows a negative deviation hence option C is correct.

The vapour pressure of the solution of two liquids $A (p^{o} = 80 m m$ ) and $B (p^{o} = 120 m m)$ is found to be $100 m m$ when $x_{A} = 0.4$ . The result shows that: