Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 5, Problem 5.10IA
Interpretation Introduction

Interpretation: The osmotic pressure of a real solution is given by ΠV=RTlnaA has to be shown.  At low concentrations the expression takes the form of ΠV=ϕRT[B] has to be shown.

Concept introduction: Osmotic pressure is the pressure that is applied to stop the flow of solvent molecules from a region of high concentration to a region of low concentration

across a semi permeable membrane.

Expert Solution & Answer
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Answer to Problem 5.10IA

The osmotic pressure of a real solution is given by ΠV=RTlnaA.  At low concentrations the expression takes the form of ΠV=ϕRT[B].

Explanation of Solution

Cell I consists of pure solvent while cell II consists of concentrated solution and osmosis occurs from cell I to cell II.

At equilibrium the fugacity of pure solvent in cell I (fsolventI) is equal to the fugacity of the solvent at cell II (fsolventII)

    fsolventI=fsolventII                                                                                                (1)

The fugacity of the solvent in the cell I is given by the expression,

    fsolvent,PIIII=xsolventIIγsolventIIfsolvent,T,PII                                                                      (2)

Where,

  • xsolventII is the mole fraction of the solvent in cell II.
  • γsolventII is the activity coefficient of the solvent.
  • fsolvent is the fugacity of the pure solvent at system temperature and pressure.
  • PII is the pressure on cell II.
  • PI is the osmotic pressure on cell I.

The fugacity of solvent at pressure PII is related to the fugacity at pressure PI by,

    fsolvent(T,PI)=fsolvent(T,PII)eVsolvent(PIIPI)RT                                                                  (3)

Substitute equation (2) in (3)

    fsolvent(T,PI)=xsolventIIγsolventIIfsolvent,T,PIIeVsolvent(PIIPI)RT

Substitute equation (1) in above.

    fsolvent(T,PII)=xsolventIIγsolventIIfsolvent,T,PIIeVsolvent(PIIPI)RT1=xsolventIIγsolventIIeVsolvent(PIIPI)RT                                                  (4)

The osmotic pressure is equal to the difference of pressure in both cells.

    Π=PIIPI

Substitute the value of PIIPI in equation (4).

    1=xsolventIIγsolventIIeVsolventΠRTΠ=RTVsolventlnxsolventIIγsolventII                                                                            (5)

If the solvent is represented by a and solute by b, then equation (5) can be written as,

    Π=RTVsolventlnxaγa                                                                                       (6)

The activity of a solvent (aa) is given by the expression,

    aa=xaγa

Substitute the value of xaγa in equation (6).

    Π=RTVsolventlnaa                                                                                            (7)

Multiply and divide equation (7) by xaxb.

    Π=xaxbxaxb×RTVsolventlnaa=xaxblnaaRTVsolvent×xbxa                                                                              (8)

Where,

  • xb is the mole fraction of the solute.

Let,

    xaxblnaa=ϕ

Where,

  • φ is the osmotic pressure coefficient.

Substitute the value of xaxblnaa in equation (8).

    ΠVsolvent=φRTxbxa                                                                                         (9)

The value of mole fraction of solute xa in dilute solutions is equal to 1.

    xa1

Substitute the value of xa in equation (9).

    ΠVsolvent=φRTxb                                                                                         (10)

The mole fraction of the solute, xb is given by the formula,

    xb=nbna+nb

Where,

  • na is the number of moles of solvent.
  • nb is the number of moles if the solute.

The number of moles of solute in a dilute solution is very less.

Hence,

    xb=nbna

Substitute the value of xb in equation (10).

    ΠVsolvent=φRTnbnaΠVsolventna=φRTnb                                                                                      (11)

In equation (11), Vna is the total volume.

Hence, equation (11) can be written as,

    ΠV=φRTnb                                                                                         (12)

The number of moles of solute, nb is

    nb=WeightofsoluteMolarmassofsolute=[B]

Hence, equation (12) can be written as,

    ΠV=φRT[B]

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Chapter 5 Solutions

Atkins' Physical Chemistry

Ch. 5 - Prob. 5A.4DQCh. 5 - Prob. 5A.5DQCh. 5 - Prob. 5A.1AECh. 5 - Prob. 5A.1BECh. 5 - Prob. 5A.2AECh. 5 - Prob. 5A.2BECh. 5 - Prob. 5A.3AECh. 5 - Prob. 5A.3BECh. 5 - Prob. 5A.4AECh. 5 - Prob. 5A.4BECh. 5 - Prob. 5A.5AECh. 5 - Prob. 5A.5BECh. 5 - Prob. 5A.6AECh. 5 - Prob. 5A.6BECh. 5 - Prob. 5A.7AECh. 5 - Prob. 5A.7BECh. 5 - Prob. 5A.8AECh. 5 - Prob. 5A.8BECh. 5 - Prob. 5A.9AECh. 5 - Prob. 5A.9BECh. 5 - Prob. 5A.10AECh. 5 - Prob. 5A.10BECh. 5 - Prob. 5A.11AECh. 5 - Prob. 5A.11BECh. 5 - Prob. 5A.1PCh. 5 - Prob. 5A.3PCh. 5 - Prob. 5A.4PCh. 5 - Prob. 5A.5PCh. 5 - Prob. 5A.6PCh. 5 - Prob. 5A.7PCh. 5 - Prob. 5B.1DQCh. 5 - Prob. 5B.2DQCh. 5 - Prob. 5B.3DQCh. 5 - Prob. 5B.4DQCh. 5 - Prob. 5B.5DQCh. 5 - Prob. 5B.6DQCh. 5 - Prob. 5B.7DQCh. 5 - Prob. 5B.1AECh. 5 - Prob. 5B.1BECh. 5 - Prob. 5B.2AECh. 5 - Prob. 5B.2BECh. 5 - Prob. 5B.3AECh. 5 - Prob. 5B.3BECh. 5 - Prob. 5B.4AECh. 5 - Prob. 5B.4BECh. 5 - Prob. 5B.5AECh. 5 - Prob. 5B.5BECh. 5 - Prob. 5B.6AECh. 5 - Prob. 5B.6BECh. 5 - Prob. 5B.7AECh. 5 - Prob. 5B.7BECh. 5 - Prob. 5B.8AECh. 5 - Prob. 5B.8BECh. 5 - Prob. 5B.9AECh. 5 - Prob. 5B.9BECh. 5 - Prob. 5B.10AECh. 5 - Prob. 5B.10BECh. 5 - Prob. 5B.11AECh. 5 - Prob. 5B.11BECh. 5 - Prob. 5B.12AECh. 5 - Prob. 5B.12BECh. 5 - Prob. 5B.1PCh. 5 - Prob. 5B.2PCh. 5 - Prob. 5B.3PCh. 5 - Prob. 5B.4PCh. 5 - Prob. 5B.5PCh. 5 - Prob. 5B.6PCh. 5 - Prob. 5B.9PCh. 5 - Prob. 5B.11PCh. 5 - Prob. 5B.13PCh. 5 - Prob. 5C.1DQCh. 5 - Prob. 5C.2DQCh. 5 - Prob. 5C.3DQCh. 5 - Prob. 5C.1AECh. 5 - Prob. 5C.1BECh. 5 - Prob. 5C.2AECh. 5 - Prob. 5C.2BECh. 5 - Prob. 5C.3AECh. 5 - Prob. 5C.3BECh. 5 - Prob. 5C.4AECh. 5 - Prob. 5C.4BECh. 5 - Prob. 5C.1PCh. 5 - Prob. 5C.2PCh. 5 - Prob. 5C.3PCh. 5 - Prob. 5C.4PCh. 5 - Prob. 5C.5PCh. 5 - Prob. 5C.6PCh. 5 - Prob. 5C.7PCh. 5 - Prob. 5C.8PCh. 5 - Prob. 5C.9PCh. 5 - Prob. 5C.10PCh. 5 - Prob. 5D.1DQCh. 5 - Prob. 5D.2DQCh. 5 - Prob. 5D.1AECh. 5 - Prob. 5D.1BECh. 5 - Prob. 5D.2AECh. 5 - Prob. 5D.2BECh. 5 - Prob. 5D.3AECh. 5 - Prob. 5D.3BECh. 5 - Prob. 5D.4AECh. 5 - Prob. 5D.4BECh. 5 - Prob. 5D.5AECh. 5 - Prob. 5D.5BECh. 5 - Prob. 5D.6AECh. 5 - Prob. 5D.1PCh. 5 - Prob. 5D.2PCh. 5 - Prob. 5D.3PCh. 5 - Prob. 5D.4PCh. 5 - Prob. 5D.5PCh. 5 - Prob. 5D.6PCh. 5 - Prob. 5D.7PCh. 5 - Prob. 5E.1DQCh. 5 - Prob. 5E.2DQCh. 5 - Prob. 5E.3DQCh. 5 - Prob. 5E.4DQCh. 5 - Prob. 5E.1AECh. 5 - Prob. 5E.1BECh. 5 - Prob. 5E.2AECh. 5 - Prob. 5E.2BECh. 5 - Prob. 5E.3AECh. 5 - Prob. 5E.3BECh. 5 - Prob. 5E.4AECh. 5 - Prob. 5E.4BECh. 5 - Prob. 5E.5AECh. 5 - Prob. 5E.5BECh. 5 - Prob. 5E.1PCh. 5 - Prob. 5E.2PCh. 5 - Prob. 5E.3PCh. 5 - Prob. 5F.1DQCh. 5 - Prob. 5F.2DQCh. 5 - Prob. 5F.3DQCh. 5 - Prob. 5F.4DQCh. 5 - Prob. 5F.5DQCh. 5 - Prob. 5F.1AECh. 5 - Prob. 5F.1BECh. 5 - Prob. 5F.2AECh. 5 - Prob. 5F.2BECh. 5 - Prob. 5F.3AECh. 5 - Prob. 5F.3BECh. 5 - Prob. 5F.4AECh. 5 - Prob. 5F.4BECh. 5 - Prob. 5F.5AECh. 5 - Prob. 5F.5BECh. 5 - Prob. 5F.6AECh. 5 - Prob. 5F.6BECh. 5 - Prob. 5F.7AECh. 5 - Prob. 5F.7BECh. 5 - Prob. 5F.8AECh. 5 - Prob. 5F.8BECh. 5 - Prob. 5F.1PCh. 5 - Prob. 5F.2PCh. 5 - Prob. 5F.3PCh. 5 - Prob. 5F.4PCh. 5 - Prob. 5.1IACh. 5 - Prob. 5.2IACh. 5 - Prob. 5.3IACh. 5 - Prob. 5.4IACh. 5 - Prob. 5.5IACh. 5 - Prob. 5.6IACh. 5 - Prob. 5.8IACh. 5 - Prob. 5.9IACh. 5 - Prob. 5.10IA
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