Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198727873
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
Question
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Chapter 4, Problem 4.1PR

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of water at 308K has to be calculated.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  lnP2P1=ΔHvapR(1T2-1T1)

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

P1= Vapor pressure at temperature T1K

P2= Vapor pressure at temperature T2K

ΔHvap= Enthalpy of vaporization

R= Universal gas constant

(a)

Expert Solution
Check Mark

Answer to Problem 4.1PR

Pressure at 308K is 0.063atm.

Explanation of Solution

According to Clausius-Clapeyron equation,

lnP2P1=ΔHvapR(1T2-1T1)

Given that,

ΔHvap=40.656kJ/mol

It is also given that ΔHvap is not dependent on temperature.

At boiling temperature of water the pressure is 1atm.

  T1=373K(boiling temperature of water)

  P1=1atm

  R=8.314Jmol-1K-1

The another temperature at which vapor pressure has to calculated is given as T2=308K

The pressure to be calculated is P2.

Applying Clausius-Clapeyron equation,

  lnP21atm=40.656×103Jmol-18.314Jmol-1K-1(1308K-1373K)lnP2=40.656×103Jmol-18.314Jmol-1K-1(1308K-1373K)P2=0.063atm

Hence the pressure at 308K is 0.063atm.

(b)

Interpretation Introduction

Interpretation:

Derivation for the modified version of Clausius-Clapeyron equation has to be done in the given condition of ΔHvap=a+bT.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  lnP2P1=ΔHvapR(1T2-1T1)

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

P1= Vapor pressure at temperature T1K

P2= Vapor pressure at temperature T2K

ΔHvap= Enthalpy of vaporization

R= Universal gas constant

(b)

Expert Solution
Check Mark

Explanation of Solution

According to Clausius-Clapeyron equation,

  lnP2P1=ΔHvapR(1T2-1T1)

It is given that ΔHvap=a+bT

Using the method of calculus on Clausius-Clapeyron equation we get,

  dlnP=ΔHvapRT2dTdlnPdT=a+bTRT2dPP=a+bTRT2dTdPP=aRT2dT+bRTdTdPP=aRdTT2+bRdTTlnP=-aRT+bRlnT

Thus the modified version of Clausius-Clapeyron equation when ΔHvap=a+bT is derived as lnP=-aRT+bRlnT.

(c)

Interpretation Introduction

Interpretation:

With the help of modified version of Clausius-Clapeyron equation, vapor pressure of water at T=308K has to be calculated.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  lnP2P1=ΔHvapR(1T2-1T1)

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

P1= Vapor pressure at temperature T1K

P2= Vapor pressure at temperature T2K

ΔHvap= Enthalpy of vaporization

R= Universal gas constant

(c)

Expert Solution
Check Mark

Answer to Problem 4.1PR

Vapor pressure calculated at T=308K is 0.0566atm.

Explanation of Solution

According to Clausius-Clapeyron equation,

  lnP2P1=ΔHvapR(1T2-1T1)

According to modified version of Clausius-Clapeyron equation when ΔHvap=a+bT,

  lnP=-aRT+bRlnT

At boiling temperature of water the pressure is 1atm.

  T1=373K(boiling temperature of water)

P1=1atm

R=8.314Jmol-1K-1

The another temperature at which vapor pressure has to calculated is given as T2=308K

The pressure to be calculated is P2.

Applying Clausius-Clapeyron equation,

  lnP=-aRT+bRlnT

  lnP2P1=-aR(1T2-1T1)+bRlnT2T1lnP21atm=-57.373kJmol-18.314Jmol-1K-1(1308K-1373K)+(-44.801kJmol-1)8.314Jmol-1K-1ln308K373KlnP2=-57.373×103Jmol-18.314Jmol-1K-1(1308K-1373K)+(-44.801×103Jmol-1)8.314Jmol-1K-1ln308K373KP2=0.0566atm

With the help of the modified Clausius-Clapeyron equation the vapor pressure calculated at T=308K is 0.0566atm.

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Chapter 4 Solutions

Elements Of Physical Chemistry

Ch. 4 - Prob. 4D.1STCh. 4 - Prob. 4D.2STCh. 4 - Prob. 4D.3STCh. 4 - Prob. 4D.4STCh. 4 - Prob. 4E.1STCh. 4 - Prob. 4E.2STCh. 4 - Prob. 4F.1STCh. 4 - Prob. 4A.1ECh. 4 - Prob. 4A.2ECh. 4 - Prob. 4A.3ECh. 4 - Prob. 4A.4ECh. 4 - Prob. 4A.5ECh. 4 - Prob. 4A.6ECh. 4 - Prob. 4B.1ECh. 4 - Prob. 4B.2ECh. 4 - Prob. 4B.3ECh. 4 - Prob. 4B.4ECh. 4 - Prob. 4B.5ECh. 4 - Prob. 4B.6ECh. 4 - Prob. 4B.7ECh. 4 - Prob. 4B.8ECh. 4 - Prob. 4C.1ECh. 4 - Prob. 4C.2ECh. 4 - Prob. 4C.3ECh. 4 - Prob. 4C.4ECh. 4 - Prob. 4C.5ECh. 4 - Prob. 4C.6ECh. 4 - Prob. 4C.7ECh. 4 - Prob. 4D.1ECh. 4 - Prob. 4D.2ECh. 4 - Prob. 4D.3ECh. 4 - Prob. 4D.4ECh. 4 - Prob. 4D.5ECh. 4 - Prob. 4D.6ECh. 4 - Prob. 4D.7ECh. 4 - Prob. 4D.8ECh. 4 - Prob. 4D.9ECh. 4 - Prob. 4D.10ECh. 4 - Prob. 4D.11ECh. 4 - Prob. 4D.12ECh. 4 - Prob. 4D.13ECh. 4 - Prob. 4E.1ECh. 4 - Prob. 4E.2ECh. 4 - Prob. 4E.3ECh. 4 - Prob. 4E.4ECh. 4 - Prob. 4F.1ECh. 4 - Prob. 4F.2ECh. 4 - Prob. 4F.3ECh. 4 - Prob. 4.1DQCh. 4 - Prob. 4.2DQCh. 4 - Prob. 4.3DQCh. 4 - Prob. 4.4DQCh. 4 - Prob. 4.5DQCh. 4 - Prob. 4.6DQCh. 4 - Prob. 4.8DQCh. 4 - Prob. 4.9DQCh. 4 - Prob. 4.10DQCh. 4 - Prob. 4.11DQCh. 4 - Prob. 4.12DQCh. 4 - Prob. 4.13DQCh. 4 - Prob. 4.14DQCh. 4 - Prob. 4.15DQCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.1PRCh. 4 - Prob. 4.2PRCh. 4 - Prob. 4.3PRCh. 4 - Prob. 4.4PRCh. 4 - Prob. 4.5PRCh. 4 - Prob. 4.6PR
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