The equation for the straight line using the given data has to be written. Concept Introduction: Clausius-Clapeyron equation: ln P = − ( Δ vap H 0 RT ) + C From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values. If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by ln P 2 p 1 = - Δ vap H 0 R [ 1 T 2 - 1 T 1 ] Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same. Normal boiling point: When the external pressure is 760 mm Hg we can call it as normal boiling point.

BuyFind

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
BuyFind

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Solutions

Chapter 11, Problem 66SCQ

(a)

Interpretation Introduction

Interpretation:

The equation for the straight line using the given data has to be written.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R[1T2-1T1]

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

(b)

Interpretation Introduction

Interpretation:

The way by which enthalpy of vaporization can be calculated using the graph has to be given.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R[1T2-1T1]

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

(c)

Interpretation Introduction

Interpretation:

The vapour pressure of ethanol at 0°Cand10°C has to be calculated.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R[1T2-1T1]

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

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