Chapter 10, Problem 60A

(a)

Interpretation Introduction

Interpretation: The value of $\Delta \text{E}$ for the process which has q = -47 kJ and w = +88 kJ needs to be calculated.

Concept Introduction: The heat flow or work indicates the energy transfer on or by the system in chemical thermodynamic.

The heat energy ‘q’ and work ‘w’ are related with total change in internal energy of system as given below:

  $\Delta \text{E = q + w}$

Answer to Problem 60A

  $\Delta \text{E = 41 kJ}$

Explanation of Solution

  $\Delta \text{E = q + w}$

q = -47 kJ

w = +88 kJ

Substitute the values:

  $\begin{array}{l}\Delta \text{E = q + w}\\ \Delta \text{E = -47 kJ + 88 kJ}\\ \Delta \text{E = 41 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta \text{E}$ for the process which has q = 82 kJ and w = +47 kJ needs to be determined.

Concept Introduction: The heat flow or work indicates the energy transfer on or by the system in chemical thermodynamic.

The heat energy ‘q’ and work ‘w’ are related with total change in internal energy of system as given below:

  $\Delta \text{E = q + w}$

Answer to Problem 60A

  $\Delta \text{E = 129 kJ}$

Explanation of Solution

  $\Delta \text{E = q + w}$

q = 82 kJ

w = +47 kJ

Substitute the values:

  $\begin{array}{l}\Delta \text{E = q + w}\\ \Delta \text{E = 82 kJ + 47 kJ}\\ \Delta \text{E = 129 kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta \text{E}$ for the process which has q = +47 kJ and w = 0 kJ needs to be determined.

Concept Introduction: The heat flow or work indicates the energy transfer on or by the system in chemical thermodynamic.

The heat energy ‘q’ and work ‘w’ are related with total change in internal energy of system as given below:

  $\Delta \text{E = q + w}$

Answer to Problem 60A

  $\Delta \text{E = 47 kJ}$

Explanation of Solution

  $\Delta \text{E = q + w}$

q = +47 kJ

w = 0 kJ

Substitute the values:

  $\begin{array}{l}\Delta \text{E = q + w}\\ \Delta \text{E = 47kJ + 0 kJ}\\ \Delta \text{E = 47 kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation: The case in which surroundings work on the system needs to be identified

1. q = -47 kJ and w = +88 kJ
2. q = 82 kJ and w = +47 kJ
3. q = +47 kJ and w = 0 kJ.

Concept Introduction: The heat flow or work indicates the energy transfer on or by the system in chemical thermodynamic.

The heat energy ‘q’ and work ‘w’ are related with total change in internal energy of system as given below:

  $\Delta \text{E = q + w}$

Answer to Problem 60A

The case with q = 82 kJ and w = +47 kJ shows the work done on the system.

Explanation of Solution

When the value of w is positive, it shows that surroundings work on the system, which makes the work done positive.

In the given three cases, (1) case has negative value of ‘w’ which indicates the work done by the system whereas (2) case has positive value of ‘w’ which indicates the work done on the system.