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.13AE

(i)

Interpretation Introduction

Interpretation: The enthalpy of vaporization of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    ln(p2p1)=ΔHvaporizationR(1T21T1)

(i)

Expert Solution
Check Mark

Answer to Problem 4B.13AE

The enthalpy of vaporization of naphthalene is 58322.71Jmol-1_.

Explanation of Solution

The enthalpy of vaporization is calculated by the Clausius-Clapeyron’s formula.

    ln(p2p1)=ΔHvaporizationR(1T21T1)                                                                 (1)

Where,

  • p1 and p2 are the pressures.
  • T1 and T2 are the temperatures.
  • ΔHvaporization is the enthalpy of vaporization of naphthalene.
  • R is the gas constant.

It is given that the temperature (T1) is 85.8°C.

The conversion of °C to K is done as,

    1°C=273K

Therefore, the conversion of 85.8°C to K is done as,

    85.8°C=85.8+273K=358.8K

Therefore, the temperature (T1) is 358.8K.

It is given that the temperature (T2) is 119.3°C.

The conversion of °C to K is done as,

    1°C=273K

Therefore, the conversion of 119.3°C to K is done as,

    119.3°C=119.3+273K=392.3K

Therefore, the temperature (T2) is 392.3K.

It is given that the pressure (p1) is 1.3kPa.

The conversion of kPa to Pa is done as,

    1kPa=1×103Pa

Therefore, the conversion of 1.3kPa to Pa is done as,

    1.3kPa=1.3×103Pa=1300Pa

Therefore, the pressure (p1) is 1300Pa.

It is given that the pressure (p2) is 5.3kPa.

The conversion of kPa to Pa is done as,

    1kPa=1×103Pa

Therefore, the conversion of 5.3kPa to Pa is done as,

    5.3kPa=5.3×103Pa=5300Pa

Therefore, the pressure (p2) is 5300Pa.

The gas constant is 8.314JK-1mol-1.

Substitute the values of p1,p2,T1,T2 and R in the equation (1).

    ln(5300Pa1300Pa)=-ΔHvaporization8.314JK-1mol-1(1392.3K-1358.8K)ln(4.07)=-ΔHvaporization8.314JK-1mol-1(0.0025K1-0.0027K1)1.403=-ΔHvaporization8.314JK-1mol-1×(-0.0002K1)

Rearrange the above equation as,

    ΔHvaporization=-1.403×8.314JK-1mol-1-0.0002K1=58322.71Jmol-1_

Hence, the enthalpy of vaporization of naphthalene is 58322.71Jmol-1_.

(ii)

Interpretation Introduction

Interpretation: The normal boiling point of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    ln(p2p1)=ΔHvaporizationR(1T21T1)

(ii)

Expert Solution
Check Mark

Answer to Problem 4B.13AE

The normal boiling point of naphthalene is 480.77K_.

Explanation of Solution

For the normal boiling point of naphthalene pressure (p2) is 1.013×105Pa.

The temperature (T1) is 358.8K.

The pressure (p1) is 1300Pa.

The enthalpy of vaporization of naphthalene is 58322.71Jmol-1.

Substitute the values of p1,p2,T1,ΔHvaporization and R in the equation (1).

    ln(1.013×105Pa1300Pa)=-58322.71Jmol-18.314JK-1mol-1(1T2-1358.8K)ln(77.92)=-7015K×(1T2-0.0027K1)4.355=-7015K×(1T2-0.0027K1)

Rearrange the above equation as,

    (1T2-0.0027K1)=4.355-7015K(1T2-0.0027K1)=0.00062K11T2=0.00062K1+0.0027K1=2.08×103K1

Reverse the value of 1T2.

    T2=12.08×103K1=480.77K_

Hence, the normal boiling point of naphthalene is 480.77K_.

(iii)

Interpretation Introduction

Interpretation: The entropy of vaporization at the boiling point of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    ln(p2p1)=ΔHvaporizationR(1T21T1)

(iii)

Expert Solution
Check Mark

Answer to Problem 4B.13AE

The entropy of vaporization at the boiling point of naphthalene is 121.31JK-1mol-1_.

Explanation of Solution

The entropy of vaporization at the boiling point of naphthalene is calculated by the following formula.

    ΔSvaporization=ΔHvaporizationTb                                                                               (2)

Where,

  • ΔSvaporization is the entropy of vaporization at the boiling point.
  • ΔHvaporization is the enthalpy of vaporization.
  • Tb is the boiling point of naphthalene.

The enthalpy of vaporization of naphthalene is 58322.71Jmol-1.

The boiling point of naphthalene is 480.77K.

Substitute the value of enthalpy of vaporization and boiling point of naphthalene in equation (4).

    ΔSvaporization=58322.71Jmol-1480.77K=121.31JK-1mol-1_

Hence, the value of the entropy of vaporization at the boiling point of naphthalene is 121.31JK-1mol-1_.

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