(a) Interpretation: Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure. Concept introduction: The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

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Answer 1

Question

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

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Answer 2

Question

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

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Answer 3

Question

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

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Answer 4

Question

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

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Answer 5

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

Answer 6

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Answer 7

Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer 8

Expert Solution

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, Tb=a/bR = 711.04 K

Molar volume for krypton υ=RT/p = 59.1 L

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT …(1)

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = 2.318atmL2/mol2

b=0.03978 L/mol

Boyle temperature Tb = a/bR

= (2.318 atm L2mol−2)/(0.03978 L mol−1x 0.08205 L. atm K−1mol−1

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

υ¯ = RT/p

= (0.08314 L bar K−1mol−1x 1 mol x 711.04 K)/1bar = 59.1 L

Answer 13

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Answer 14

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Answer 15

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer 16

Expert Solution

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The non-ideal gas equation represented as;

[p+an2/V2][V–nb]=nRT…(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

=(5.489 atm L2mol−2)/(0.0638 L mol−1x 0.08205 L. atm K−1mol−1= 1049.5 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 1049.5 K)/1bar= 87.2 L

Answer 21

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Answer 22

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

Answer 23

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer 24

Expert Solution

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The non-ideal gas equation represented as;

[p + an2/V2] [ V – nb] = nRT …(1)

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

=(0.08314 L bar K−1mol−1x 1 mol x 5822 K)/1bar= 484 L

Answer 29

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.

Answer 30

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Answer to Problem 1.79E

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Answer to Problem 1.79E

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