To prove that T ( ∂ Z ∂ T ) ρ = ρ ( ∂ Z ∂ ρ ) T and apply it to Van der Waals equation Concept Introduction: Using, P = ρ Z R T to prove T ( ∂ Z ∂ T ) ρ = ρ ( ∂ Z ∂ ρ ) T By Joule Thompson inversion curve, ( ∂ Z ∂ T ) P = 0 Then applying it to General equation which is given as, P = R T V − b − a ( T ) ( V + ∈ b ) × ( V + σ b )

Question

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Answer 1

Question

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

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Answer 2

Question

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

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Answer 3

Question

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

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Answer 4

Question

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

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Answer 5

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 6

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 7

Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Van der Waals equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

Answer 8

(a)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ (A)

ξ=1−12q (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+σb)

With Van der Waals parameters becomes:

P=RTV−b−aV2

Multiply the above equation by V/RT,

PVRT=VV−b−aVRT

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−aρRT

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ (A)

Differentiating the above equation, we get

(∂Z∂T)ρ=(∂Z∂T)ξ=−ξdqdT

By q=ψa(Tr)ΩTr with α(Tr)=1 for the van der waals equation, q=ψΩTr We get,

dqdT=ψΩ(−1Tr2)dTrdT=−ψΩ1Tr2Tc=−ψΩ1TTr=−qT

Then,

(∂Z∂T)ρ=(−ξ)(−qT)=qξT

In addition,

(∂Z∂ρ)T=b(∂Z∂ξ)T=b(1−ξ)2−qb

Substitute for the two partial derivatives in the boxed equation:

TqξT=bρ(1−ξ)2−qbρ or

qξ=ξ(1−ξ)2−qξ

hence,

ξ=1−12q (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 13

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 14

(b)

Interpretation Introduction

Interpretation:

To prove that T(∂Z∂T)ρ=ρ(∂Z∂ρ)T and apply it to Redlich Kwong equation

Concept Introduction:

Using, P=ρZRT to prove T(∂Z∂T)ρ=ρ(∂Z∂ρ)T

By Joule Thompson inversion curve, (∂Z∂T)P=0

Then applying it to General equation which is given as,

P=RTV−b−a(T)(V+∈b)×(V+σb)

Answer 15

(b)

Expert Solution

Answer to Problem 7.13P

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

Z=11−ξ−qξ1+ξ (A)

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

(∂x∂y)z=(∂x∂y)w+(∂x∂w)y(∂w∂y)z

(∂Z∂T)p=(∂Z∂T)ρ+(∂Z∂p)T(∂ρ∂T)P

We get,(∂Z∂T)ρ=(∂Z∂T)P−(∂Z∂ρ)p(∂ρ∂T)P

Now, P=ρZRT i.e. ρ=PZRT

(∂ρ∂T)P=PR{−1( ZT)2[Z+T( ∂Z ∂T)P]}

Substituting (∂Z∂T)P=0 in each of the two preceding equations reduces them to:

(∂Z∂T)ρ=−(∂Z∂ρ)T(∂ρ∂T)P

(∂ρ∂T)P=−PZRT2=−ρT

Combing these two equations yields:

T(∂Z∂T)ρ=ρ(∂Z∂ρ)T Hence Proved

The general equation is as follows,

P=RTV−b−a(T)(V+∈b)×(V+b)

With Redlich Kwong parameters becomes:

P=RTV−b−aV(V+b)

Multiply the above equation by V/RT,

PVRT=VV−b−aRT(V+b)

substitute Z = PV/RT and V=1/ρ,

Z=11−bρ−abρbRT(1+bρ)

Assume, q=abRT and define ξ=bρ, we get,

Z=11−ξ−qξ1+ξ (A)

By q=ψa(Tr)ΩTr with α(Tr)=Tr−0.5 for the Redlich/Kwong equation, q=ψΩTr1.5 .

We get,

dqdT=−1.5qTand (∂Z∂T)ρ=1.5qξT(1+ξ)

(∂Z∂ρ)T=b(1−ξ)2−bq(1+ξ)2

Substitution into the above equation, we get

q=(1+ξ1−ξ)2(12.5+1.5ξ) (B)

Now, b=ΩRTCPC

Pc=ΩRTcb

(I)

P=ZρRT

(II)

Division of (II) by (I) gives

Pr=PPC=ZρbTΩTc

Pr=ZξTrΩ (C)

For a given T_r, we can find q and then use equation (B) to find ξ

The Equation (A) we can find Z and then P_r with equation (C)

Answer 20

Ch. 7 - Prob. 7.1P Ch. 7 - Prob. 7.2P Ch. 7 - Prob. 7.3P Ch. 7 - Prob. 7.4P Ch. 7 - Prob. 7.5P Ch. 7 - Prob. 7.6P Ch. 7 - Prob. 7.7P Ch. 7 - Prob. 7.8P Ch. 7 - Prob. 7.9P Ch. 7 - Prob. 7.10P

Answer to Problem 7.13P

Explanation of Solution

Answer to Problem 7.13P

Explanation of Solution

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