Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
Question
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Chapter 4, Problem 4.2PR

(a)

Interpretation Introduction

Interpretation:

The expression for the change in molar Gibbs energy has to be derived.

Concept Introduction:

The van der Waals equation of state can be expressed as

(P+aVm2)(Vmb)=RT

Where,

Vm=Vn= Molar volume

a= A constant for attractive effect

b= A constant for repulsive effect

P= Pressure

T= Absolute temperature

R= Universal gas constant

Ideal gas equation can also be represented as

PV=nRT

Molar Gibbs energy can be mathematically represented in terms of T and P,

dGm=VmdPSdT

At constant T,

dGm=VmdP

The change in molar Gibbs energy for an ideal gas is,

ΔGmIdeal=RTln(PfPi)

Where,

Pf= Final pressure

Pi= Initial pressure

(a)

Expert Solution
Check Mark

Explanation of Solution

The van der Waals equation of state can be expressed as

(P+aVm2)(Vmb)=RT

According to the give question, we can neglect attractive effects.  Then the equation becomes,

P(Vmb)=RT

Solving for Vm,

P(Vmb)=RTPVmPb=RTVm=(RTP)+b.

Where,

RTP= Molar volume of a perfect gas

By taking integration from limit initial value to final value,

GiGfdGm=PiPfVmdPGfGi=PiPf{(RTP)+b}dPΔGm=RTln(PfPi)+b(PfPi)

The expression in the above box represents the change in molar Gibbs energy for a real gas.

(b)

Interpretation Introduction

Interpretation:

Whether the change in molar Gibbs energy is greater or smaller than for a perfect gas has to be decided.

Concept Introduction:

The change in molar Gibbs energy for an ideal gas is,

ΔGmIdeal=RTln(PfPi)

The change in molar Gibbs energy for a perfect gas is,

ΔGm=RTln(PfPi)+b(PfPi)

(b)

Expert Solution
Check Mark

Answer to Problem 4.2PR

The change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor b(PfPi).

Explanation of Solution

The change in molar Gibbs energy for an ideal gas is,

ΔGmIdeal=RTln(PfPi)

The change in molar Gibbs energy for a perfect gas is,

ΔGm=RTln(PfPi)+b(PfPi)

The change,

ΔGmΔGmIdeal=[RTln(PfPi)+b(PfPi)][RTln(PfPi)]=b(PfPi)

Therefore, the change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor b(PfPi).

(c)

Interpretation Introduction

Interpretation:

The percentage difference between van der Walls and perfect gas for carbon dioxide has to be calculated.

Concept Introduction:

The change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor b(PfPi).

Percentage increase =NewnoumberOriginalnoumberOriginalnoumber×100

(c)

Expert Solution
Check Mark

Answer to Problem 4.2PR

The percentage difference has been calculated to be 0.6861%.

Explanation of Solution

Given data:

Pi=1.0atmPf=10.0atm

Calculation of percentage difference:

ΔGmIdeal=RTln(PfPi)=(0.0821L.atm.K1.mol1)(298K)ln(10.01.0)=56.27L.atm.mol1.

ΔGmΔGmIdeal=[RTln(PfPi)+b(PfPi)][RTln(PfPi)]=b(PfPi)=(4.29)(10.01.0)=38.61×102L.mol1.

Percentage difference:

ΔGmΔGmIdealΔGmIdeal×100=38.61×102L.mol156.27L.atm.mol1×100=0.6861%.

Therefore, the calculated percentage difference is 0.6861%.

(d)

Interpretation Introduction

Interpretation:

The critical constants for the given gas in terms of the parameters ‘a’ and ‘b’ has to be derived.

(d)

Expert Solution
Check Mark

Explanation of Solution

A gas described by the equation of state

P=nRTVan2V2+bn3V3........(1)

If we plot a graph between P and V which is called an isotherm, we will get a flat inflexion where dPdV=0andd2PdV2=0.  It implies that the gas described by the above equation shows critical behavior.

Calculation of critical constants:

Critical Volume(Vc):

Taking the first and second derivative and equalizing it to zero, we can get two equations,

2an2V3nRTV23bn3V4=0.........(2)6an2V4+2nRTV3+12bn3V5=0.........(3)

Consider equation (1). We can express T in terms of V as below,

T=[3bn3V42an2V3]/nR=[3bn32an2VnRV2]........(4)

Now substituting the value of T in equation (2) and solving for Vc,

Vc=3bna

Critical Temperature(Tc):

By substituting the value of Vc in equation (3) and solving for Tc,

Tc=a23bR

By substituting the value of Vc and Tc in the equation (1) and solving for Pc,

Pc=a327b2

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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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