The vapor pressure of benzene at 60 .6°C has to be calculated. Concept Introduction: Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows – ln P 1 P 2 = ΔH vap R [ 1 T 2 - 1 T 1 ] Where P 1 = vapor pressure of the liquid at temperature T 1 . P 2 = vapor pressure of the liquid at temperature T 2 . ΔH vap = Molar heat of vaporization R = Universal Gas constant .

Question

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Answer 1

Question

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

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Answer 2

Question

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

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Answer 3

Question

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

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Answer 4

Question

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

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Answer 5

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

Answer 6

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$

Conclusion

The vapor pressure of benzene at $60.6°C$ is calculated.

Answer 7

Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6°C$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$ln P1P2 = ΔHvap R [1T2 - 1T1]Where P1 = vapor pressure of the liquid at temperature T1. P2 = vapor pressure of the liquid at temperature T2. ΔHvap = Molar heat of vaporization R = Universal Gas constant.$

Answer 8

Expert Solution & Answer

Answer to Problem 11.84QP

The vapor pressure of benzene at $60.6°C$ is calculated as $331mmHg.$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Vapor pressure of benzene at $7.6°C$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $60.6°C$ is calculated using the formula $ln P1P2 = ΔHvap R [1T2 - 1T1]$ in which known values are substituted and the required parameter is obtained as follows –

$Given data: ΔHvap = 31.0 kJ/mol P1 = 40.1 mmHg T1 = 7.6°C = 280.8K T2 = 60.6°C = 333.8K P2 = ?$

$ln P1P2 = ΔHvap R [1T2 - 1T1]ln 40.1 mmHgP2 = 31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) 40.1 mmHgP2 = e31.0 × 103 J8.314 J/K.mol (1333.8 K - 1280.8 K) = 0.121 P2 = 40.1 mmHg0.121 = 331 mmHg$