Chapter 9, Problem 125MP

Interpretation Introduction

Interpretation: The value of q, w, $\Delta E$ and $\Delta H$ for each step needs to be calculated. Also, the overall values of each pathway need to be determined. This is to be explained from the overall change in entropy and change in enthalpy that it is a state functions and q and w are path functions.

Concept Introduction:

For a process at constant pressure, the work dome can be calculated as follows:

  $w=-P\Delta V$

Also, value of heat, q can be calculated as follows:

  $q_P=n{C}_P\Delta T$

The change in enthalpy is equal to $q_P$ for constant pressure condition. The change in internal energy can be calculated s follows:

  $\Delta E=q+w$

Here, q is heat and w is work done.

Explanation of Solution

For the first step which is at constant pressure or isobaric in nature, the work done can be calculated as follows:

  $w=-P\Delta V$

Putting the values,

  $\begin{array}{l}w=-\left(3\text{ atm}\right)\left(55-15\right)\text{L}\\ \text{=}\left(120\text{ L}\text{.atm}\right)\left(\frac{0.1013\text{ kJ}}{\text{1 L}\text{.atm}}\right)\\ =-12.16\text{ kJ}\end{array}$

For the process which is at constant pressure, the enthalpy change is equal to the value of q which is calculated as follows:

  $q_P=n{C}_P\Delta T$

This is can be represented as follows:

  $\begin{array}{l}{q}_P=n\left(\frac{5R}{2}\right)\Delta \left(\frac{PV}{nR}\right)\\ =\frac{5}{2}P\Delta V\end{array}$

Putting the values,

  $\begin{array}{l}{q}_P=\frac{5}{2}P\Delta V=\frac{5}{2}\left(3\text{ atm}\right)\left(55-15\right)\text{ L}\\ \text{=300 L}\text{.atm}\left(\frac{0.1013\text{ kJ}}{1\text{ L}\text{.atm}}\right)\\ =30.4\text{ kJ}\end{array}$

Now,

  $\Delta H=q_P=30.4\text{ kJ}$

Now, change in internal in internal energy can be calculated as follows:

  $\Delta E=q+w=\left(30.4-12.16\right)\text{kJ=17}\text{.8 kJ}$

Now, for step 2, there is no constant pressure and volume thus, value of q can be calculated using the external pressure and volume change as follows:

  $\begin{array}{l}w=-P_{ext}\Delta V\\ =-\left(6\text{ atm}\right)\left(20-55\right)\text{ L}\left(\frac{0.1013\text{ kJ}}{1\text{ L atm}}\right)\\ =21.3\text{ kJ}\end{array}$

Change in enthalpy can be calculated as follows:

  $\Delta H=n{C}_P\Delta T$

This can be rewritten as follows:

  $\begin{array}{l}\Delta H=n\frac{5R}{2}\Delta \frac{PV}{nR}\\ =\frac{5}{2}\Delta \left(PV\right)\\ =\frac{5}{2}\left(120-165\text{ L atm}\right)\left(\frac{0.1013\text{ kJ}}{\text{1 L atm}}\right)\\ =-11.4\text{ kJ}\end{array}$

Now, change in internal energy can be calculated as follows:

  $\begin{array}{l}\Delta E=n\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{ R}\right)\Delta \left(\frac{PV}{nR}\right)\\ =\frac{3}{2}\left(120-165\text{ L}\text{.atm}\right)\left(\frac{0.1013\text{ kJ}}{1\text{ L}\text{.atm}}\right)\\ =-6.8\text{ kJ}\end{array}$

Now, the value of q can be calculated as follows:

  $\begin{array}{l}q=\Delta E-w\\ \text{=}\left(\text{-6}\text{.8-21}\text{.3}\right)\text{ kJ}\\ =-28.1\text{ kJ}\end{array}$

Now, the overall values can be calculated as follows:

  $\begin{array}{l}w=-12.16+21.3=9.14\text{ kJ}\\ q=30.4-28.1=2.3\text{ kJ}\\ \Delta E=17.8-6.8=11.0\text{ kJ}\\ \Delta H=30.4-11.4=19\text{ kJ}\end{array}$

The given step 3 is an isochoric process that is the process takes place at constant volume. The value of work done is zero as there is no change in volume.

The change in enthalpy can be calculated as follows:

  $\begin{array}{l}\Delta H=\frac{5}{2}\Delta \left(PV\right)=\frac{5}{2}\left(90-45\text{ L atm}\right)\left(\frac{0.1013\text{ kJ}}{1\text{ L atm}}\right)\\ =11.4\text{ kJ}\end{array}$

Similarly, change in internal energy can be calculated as follows:

  $\begin{array}{l}\Delta H=\frac{3}{2}\Delta \left(PV\right)=\frac{3}{2}\left(90-45\text{ L atm}\right)\left(\frac{0.1013\text{ kJ}}{1\text{ L atm}}\right)\\ =6.84\text{ kJ}\end{array}$

The value of q is equal to change in internal energy thus,

  $q=\Delta E=6.84\text{ kJ}$

Step 4 is isobaric in nature thus, work done can be calculated as follows:

  $\begin{array}{l}w=-P\Delta V\\ =-\left(6\text{ atm}\right)\left(20-15\right)\text{ L}\left(\frac{0.1013\text{ kJ}}{1\text{ L atm}}\right)\\ =-3\text{ kJ}\end{array}$

The enthalpy change for an ideal gas can be calculated as follows:

  $\begin{array}{l}\Delta H=\frac{5}{2}\Delta \left(PV\right)\\ =\frac{5}{2}\left(120-90\text{ L atm}\right)\left(\frac{0.1013\text{ kJ}}{1\text{ L atm}}\right)\\ =7.6\text{ kJ}\end{array}$

The value of q is equal to change in enthalpy as follows:

  $\begin{array}{l}q=\Delta H\\ =7.6\text{ kJ}\end{array}$

The change in internal energy can be calculated as follows:

  $\begin{array}{l}\Delta E=q+w\\ =\left(7.6-3.0\right)\text{ kJ}\\ =\text{4}\text{.6 kJ}\end{array}$

The overall change can be calculated as follows:

  $\begin{array}{l}w=\left(0-3.0\right)\text{ kJ}=-3.0\text{ kJ}\\ q=\left(6.84+7.6\right)\text{ kJ}=14.4\text{ kJ}\\ \Delta E=\left(6.84+4.6\right)\text{ kJ}=11.4\text{ kJ}\\ \Delta \text{H}=\left(\text{ 11}\text{.4+7}\text{.6}\right)\text{ kJ=19}\text{.0 kJ}\end{array}$

Conclusion

On comparing the overall values, change in internal energy and enthalpy can be seen same for both the paths.

Thus, the internal and enthalpy change are path independent or they are state functions.

Also, the values of q and w are different for two paths thus, they are path functions.