PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH)
PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH)
11th Edition
ISBN: 9780198826910
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 4, Problem 4B.4P

(a)

Interpretation Introduction

Interpretation: The value of dp/dT using the Clapeyron equation at the normal boiling point has to be estimated.

Concept introduction: The value of dp/dT using the Clapeyron equation is calculated by using the Clapeyron’s equation.  The Clapeyron’s equation is shown below.

    (dpdT)=ΔvapHTΔVm

(a)

Expert Solution
Check Mark

Answer to Problem 4B.4P

The value of dp/dT using the Clapeyron equation at the normal boiling point is 0.0055PaK-1_.

Explanation of Solution

The Clapeyron equation is shown below.

    dpdT=ΔvapHTΔVm                                                                                                  (1)

Where,

  • dpdT is the change in pressure with respect to change in temperature.
  • ΔvapH is the enthalpy of vaporization.
  • T is the normal boiling point of the liquid.
  • ΔVm is the change in molar volume.

It is given that the enthalpy of vaporization is 14.4kJmol-1.

The conversion of kJmol-1 to Jmol-1 is done as,

    1kJmol-1=1×103Jmol-1

Therefore, the conversion of 14.4kJmol-1 to Jmol-1 is done as,

    14.4kJmol-1=14.4×103Jmol-1

Therefore, the enthalpy of vaporization of liquid is 14.4×103Jmol-1.

It is given that the normal boiling point of the liquid is 180K.

It is given that the molar volume of the liquid (Vm1) and the vapour state (Vm2) is 115cm3mol-1 and 14.5dm3mol-1 respectively.

The conversion of cm3mol-1 to m3mol-1 is done as,

    1cm3mol-1=1×106m3mol-1

Therefore, the conversion of 115cm3mol-1 to m3mol-1 is done as,

    115cm3mol-1=115×106m3mol-1

Therefore, the molar volume in the liquid state (Vm1) is 115×106m3mol-1.

The conversion of dm3mol-1 to m3mol-1 is done as,

    1dm3mol-1=1×103m3mol-1

Therefore, the conversion of 14.5dm3mol-1 to m3mol-1 is done as,

    14.5dm3mol-1=14.5×103m3mol-1

Therefore, the molar volume in the liquid state (Vm2) is 14.5×103m3mol-1.

The change in molar volume is calculated by the following formula.

    ΔVm=Vm2Vm1                                                                                             (2)

Substitute the value of the molar volume in liquid and vapour state.

    ΔVm=14.5×103m3mol-1115×106m3mol-1=14.5×103m3mol-10.115×103m3mol-1=14.385×103m3mol-1

Therefore, the change in molar volume (ΔVm) is 14.385×103m3mol-1.

Substitute the value of ΔvapH,ΔVm and T in equation (1).

    dpdT=14.4×103Jmol-1180K×14.385×103m3mol-1=0.0055Jm-3K-1

The value of dp/dT using the Clapeyron equation at the normal boiling point is 0.0055Jm-3K-1.

The conversion of Jm-3K1 to PaK1 is done as,

    1Jm-3K1=1PaK1

Therefore the conversion of 0.0055Jm-3K-1 to PaK1 is done as,

    0.0055Jm-3K-1=0.0055PaK-1_

Hence, the value of dp/dT using the Clapeyron equation at the normal boiling point is 0.0055PaK-1_.

(b)

Interpretation Introduction

Interpretation: The percentage error in the resulting value of dp/dT has to be estimated.

Concept introduction: The Clausius-Clapeyron’s equation shows the correlation between the temperature and pressure.  The Clausius-Clapeyron’s equation can be used for the phase transition process.  The Clausius-Clapeyron’s equation is shown below.

    dpdT=pΔvapHRT2

(b)

Expert Solution
Check Mark

Answer to Problem 4B.4P

The percentage error in the resulting value of dp/dT is 1.63%_.

Explanation of Solution

The value of dp/dT is calculated by the Clausius-Clapeyron’s equation.

    dpdT=pΔvapHRT2                                                                                               (3)

Where,

  • ΔvapH is the enthalpy of vaporization.
  • T is the normal boiling point.
  • R is the gas constant.

Substitute the value of p,ΔvapH,R and T in equation (3).

    dpdT=1.013×105Pa×14.4×103Jmol-18.314JK-1mol-1×(180K)2=14.58×102PaJmol-1269373.6JKmol-1=0.00541PaK1

The percentage error in the resulting value of dp/dT is calculated by the following formula.

    %Error=((dp/dT)Clapeyron's(dp/dT)Clausius-Clapeyron(dp/dT)Clapeyron's)×100%                        (4)

The value of dp/dT from Clapeyron equation is 0.0055PaK-1.

The value of dp/dT from Clausius-Clapeyron equation is 0.0055PaK-1.

Substitute the values of dp/dT in equation (4).

    %Error=(0.0055PaK-10.00541PaK10.0055PaK-1)×100%=1.63%_

Hence, the percentage error in the resulting value of dp/dT is 1.63%_.

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