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.

Question

Want to see the full answer?

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 11

Ch. 9 - Prob. 1QAP Ch. 9 - Prob. 2QAP Ch. 9 - The vapor pressure of I2(s) at 30C is 0.466 mm Hg....Ch. 9 - Prob. 4QAP Ch. 9 - Trichloroethane, C2H3Cl3 is used as a degreaser...Ch. 9 - p-Dichlorobenzene, C6H4Cl2, can be one of the...Ch. 9 - Prob. 7QAP Ch. 9 - Dichloromethane, CH2Cl2,is widely used as a...Ch. 9 - Prob. 9QAP Ch. 9 - Prob. 10QAP

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.

Want to see the full answer?

Chapter 9 Solutions