PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
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Chapter 15, Problem 15A.4P
Interpretation Introduction

Interpretation:

The expressions for the volume of a triclinic, monoclinic, orthorhombic and tetragonal unit cells have to be derived and also the volume of a given tetragonal unit cell has to be calculated.

Concept Introduction:

From a vector analysis, we know that volume of a geometric system can be alternatively written as

  V=a.b×c

Expert Solution & Answer
Check Mark

Answer to Problem 15A.4P

The volume of the tetragonal unit cell is found out to be 395830134×10-24m3.

Explanation of Solution

First we have to derive the volume of a triclinic unit cell. Then we can take it as the reference to derive the volume of a hexagonal unit cell.

Derivation of the volume of a triclinic unit cell:

The cell dimensions of a triclinic unit cell is abcandα=β=γ=90.

PHYSICAL CHEMISTRY-WEBASSIGN, Chapter 15, Problem 15A.4P

Figure 1

There are many choices for the primitive vectors in the triclinic system. So we are taking

  a=axi^+ayj^+azk^b=bxi^+byj^+bzk^c=cxi^+cyj^+czk^

Then,

  V=a.b×c=axayazbxbybzcxcycz

Therefore,

  V2=axayazbxbybzcxcyczaxayazbxbybzcxcycz=axayazbxbybzcxcyczaxbxcxaybycyazbzcz[Interchangerowsandcolumns]=axax+ayay+azazaxbx+ayby+azbzaxcx+aycy+azczbxax+byay+bzazbxbx+byby+bzbzbxcx+bycy+bzczcxax+cyay+czazcxbx+cyby+czbzcxcx+cycy+czcz=a2a.ba.cb.ab2b.cc.ac.bc2=a2abcosγaccosβabcosγb2bccosαaccosβbccosαc2=a2b2c2(1cos2αcos2βcos2γ+2cosαcosβcosγ)Hence,V=abc(1cos2αcos2βcos2γ+2cosαcosβcosγ)1/2

This is the expression for the volume of a triclinic unit cell.

Derivation of the volume of a monoclinic unit cell:

The cell dimensions of a monoclinic unit cell is abcandα=γ=90β.

Now substituting these values in the expression of the volume of a triclinic unit cell, we can get the exact expression for the monoclinic unit cell.

  V=abc(1cos2αcos2βcos2γ+2cosαcosβcosγ)1/2=abc[1cos2(90)cos2βcos2(90)+2cos(90)cosβcos(90)]1/2=abc[1cos2β]1/2=abcsinβ.[cos(90)=01cos2β=sin2β]

The above expression is for the volume of a monoclinic unit cell.

Derivation of the volume of an orthorhombic unit cell:

The cell dimensions of a orthorhombic unit cell is abcandα=γ=β=90.

Now substituting these values in the expression of the volume of a triclinic unit cell, we can get the exact expression for the orthorhombic unit cell.

V=abc(1cos2αcos2βcos2γ+2cosαcosβcosγ)1/2=abc[1cos2(90)cos2(90)cos2(90)+2cos(90)cos(90)cos(90)]1/2=abc[cos(90)=0]

The above expression is for the volume of an orthorhombic unit cell.

Derivation of the volume of a tetragonal unit cell:

The cell dimensions of a tetragonal unit cell is a=bcandα=γ=β=90.

Now substituting these values in the expression of the volume of a triclinic unit cell, we can get the exact expression for the tetragonal unit cell.

V=abc(1cos2αcos2βcos2γ+2cosαcosβcosγ)1/2=a.a.c[1cos2(90)cos2(90)cos2(90)+2cos(90)cos(90)cos(90)]1/2=a2c[cos(90)=0]

The above expression is for the volume of a tetragonal unit cell.

Given data:

For a tetragonal unit cell, a=651pmandc=934pm.

The expression for the volume of a tetragonal unit cell is V=a2c.

Hence,

V=a2c=(651pm)2(934pm)=395830134×1024m3.

Therefore, the volume of the tetragonal unit cell is 395830134×1024m3.

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Chapter 15 Solutions

PHYSICAL CHEMISTRY-WEBASSIGN

Ch. 15 - Prob. 15A.2AECh. 15 - Prob. 15A.2BECh. 15 - Prob. 15A.3AECh. 15 - Prob. 15A.3BECh. 15 - Prob. 15A.4AECh. 15 - Prob. 15A.4BECh. 15 - Prob. 15A.1PCh. 15 - Prob. 15A.2PCh. 15 - Prob. 15A.3PCh. 15 - Prob. 15A.4PCh. 15 - Prob. 15A.5PCh. 15 - Prob. 15A.6PCh. 15 - Prob. 15A.7PCh. 15 - Prob. 15A.8PCh. 15 - Prob. 15A.9PCh. 15 - Prob. 15B.1DQCh. 15 - Prob. 15B.2DQCh. 15 - Prob. 15B.3DQCh. 15 - Prob. 15B.1AECh. 15 - Prob. 15B.1BECh. 15 - Prob. 15B.2AECh. 15 - Prob. 15B.2BECh. 15 - Prob. 15B.3AECh. 15 - Prob. 15B.3BECh. 15 - Prob. 15B.4AECh. 15 - Prob. 15B.4BECh. 15 - Prob. 15B.5AECh. 15 - Prob. 15B.5BECh. 15 - Prob. 15B.6AECh. 15 - Prob. 15B.6BECh. 15 - Prob. 15B.7AECh. 15 - Prob. 15B.7BECh. 15 - Prob. 15B.11AECh. 15 - Prob. 15B.11BECh. 15 - Prob. 15B.12AECh. 15 - Prob. 15B.12BECh. 15 - Prob. 15B.1PCh. 15 - Prob. 15B.2PCh. 15 - Prob. 15B.3PCh. 15 - Prob. 15B.4PCh. 15 - Prob. 15B.6PCh. 15 - Prob. 15B.7PCh. 15 - Prob. 15C.1DQCh. 15 - Prob. 15C.2DQCh. 15 - Prob. 15C.1AECh. 15 - Prob. 15C.2AECh. 15 - Prob. 15C.2BECh. 15 - Prob. 15C.3AECh. 15 - Prob. 15C.3BECh. 15 - Prob. 15C.4AECh. 15 - Prob. 15C.4BECh. 15 - Prob. 15C.5AECh. 15 - Prob. 15C.5BECh. 15 - Prob. 15C.1PCh. 15 - Prob. 15C.2PCh. 15 - Prob. 15C.3PCh. 15 - Prob. 15C.4PCh. 15 - Prob. 15C.5PCh. 15 - Prob. 15C.7PCh. 15 - Prob. 15C.8PCh. 15 - Prob. 15C.9PCh. 15 - Prob. 15D.1DQCh. 15 - Prob. 15D.1AECh. 15 - Prob. 15D.1BECh. 15 - Prob. 15D.2AECh. 15 - Prob. 15D.2BECh. 15 - Prob. 15D.3AECh. 15 - Prob. 15D.3BECh. 15 - Prob. 15D.1PCh. 15 - Prob. 15D.2PCh. 15 - Prob. 15E.1DQCh. 15 - Prob. 15E.1AECh. 15 - Prob. 15E.1BECh. 15 - Prob. 15E.2AECh. 15 - Prob. 15E.2BECh. 15 - Prob. 15E.3AECh. 15 - Prob. 15E.3BECh. 15 - Prob. 15E.5PCh. 15 - Prob. 15F.1DQCh. 15 - Prob. 15F.1AECh. 15 - Prob. 15F.1BECh. 15 - Prob. 15F.2AECh. 15 - Prob. 15F.2BECh. 15 - Prob. 15F.3AECh. 15 - Prob. 15F.3BECh. 15 - Prob. 15F.4AECh. 15 - Prob. 15F.4BECh. 15 - Prob. 15F.5AECh. 15 - Prob. 15F.5BECh. 15 - Prob. 15G.1DQCh. 15 - Prob. 15G.2DQCh. 15 - Prob. 15G.1AECh. 15 - Prob. 15G.1BECh. 15 - Prob. 15.1IA
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