The vapor pressure of octane is obtained by the following data- Vp(mmHg) 10 40 100 400 t (°C) 19.2 45.1 65.7 104.0 The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph. Concept introduction: The relation between pressure, temperature and change in enthalpy of vaporization is as follows: lnP = A − Δ H v a p R ( 1 T ) Here, P = pressure T = temperature Δ H v a p = heat of vaporization = slope A = y-intercept from the graph.

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Answer 1

Question

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

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Answer 2

Question

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Expert Solution & Answer

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Answer 4

Question

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Expert Solution & Answer

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Answer 5

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Answer 6

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Answer 7

Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg) 10 40 100 400

t (°C) 19.2 45.1 65.7 104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

lnP=A−ΔHvapR(1T)

Here,

P = pressure

T = temperature

ΔHvap = heat of vaporization = slope

A = y-intercept from the graph.

Answer 8

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Answer 9

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Answer 10

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Answer 11

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Solution Summary: The author explains how the graph between ln(vp) and 1/T is to be plotted and the heat of vaporization is estimated.

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Chapter 9 Solutions