The standard molar entropy values of N e a n d X e has to be compared. Concept Introduction: The entropy of distinguishable particle in terms of partition function can be calculated using following formula. S = U − U ( 0 ) T + N k ln q The entropy of indistinguishable particle in terms of partition function can be calculated using following formula. S = U − U ( 0 ) T + N k ln q − N k ( ln N − 1 ) Where, S = e n t r o p y U − U ( 0 ) = c h a n g e i n int e r n a l e n e r g y k = p l a n k ' s c o n s tan t q = p a r t i t i o n f u n c t i o n So from the above equation it is clear that the entropy depends on partition functions . The expression for the translational partition function is given below. q T = ( 2 π m k T ) 3 2 V h 3 Where, q T = t r a n s l a t i o n a l p a r t i t i o n f u n c t i o n m = m o l e c u l a r m a s s T = t e m p e r a t u r e h = p l a n k ' s c o n s tan t V = v o l u m e a t s tan d a r d c o n d i t i o n s k = b o l t z m a n n c o n s tan t The expression for rotational partition function is given below. q R = k T σ h B Where, q R = r o t a t i o n a l p a r t i t i o n f u n c t i o n k = b o l t z m a n n c o n s tan t T = t e m p e r a t u r e h = p l a n k ' s c o n s tan t σ = s y m m e t r y n u m b e r B = r o t a t i o n a l c o n s tan t The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

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Answer 1

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

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Answer 2

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

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Answer 3

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 6

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 7

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 8

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 9

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 10

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 11

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 12 - Prob. 12A.1ST Ch. 12 - Prob. 12A.2ST Ch. 12 - Prob. 12B.2ST Ch. 12 - Prob. 12B.3ST Ch. 12 - Prob. 12B.5ST Ch. 12 - Prob. 12C.1ST Ch. 12 - Prob. 12C.3ST Ch. 12 - Prob. 12C.4ST Ch. 12 - Prob. 12A.1E Ch. 12 - Prob. 12A.2E

Videos

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Want to see the full answer?

Chapter 12 Solutions

Videos

Interpretation: The standard molar entropy values of C(diamond) and C(graphite) has to be compared. Concept Introduction: The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Want to see the full answer?

Chapter 12 Solutions

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.