Chapter 10, Problem 147CP

Consider 1.00 mole of an ideal gas that is expanded isothermally at 25°C from $2.45\times {10}^{-2}$ atm to $2.45\times {10}^{-3}$ atm in the following three irreversible steps:
Step 1: from $\text{2}{\text{.45×10}}^{\text{-2}}\text{atm to 9}{\text{.87×10}}^{\text{-3}}\text{atm}$
Step 2: from $\text{9}{\text{.87×10}}^{\text{-3}}\text{atm to 4}{\text{.93×10}}^{\text{-3}}\text{atm}$
Step 3: from $\text{4}{\text{.93×10}}^{\text{-3}}\text{atm to 2}{\text{.45×10}}^{\text{-3}}\text{atm}$
Calculate q, w, $\Delta E\text{,}\Delta S\text{,}\Delta H\text{, and}\Delta G$ for each step andfor the overall process.

Interpretation Introduction

Interpretation : The value of $q,w,\Delta E,\Delta S,\Delta H\text{ and }\Delta G$ for each step and the overall process should be calculated.

Concept Introduction :

Work done can be calculated as follows:

  $w=-P.\Delta V$

  $\Delta V=V_f-V_i=nRT\left(\frac{1}{P_f}-\frac{1}{P_i}\right)$

The internal energy is sum of heat and work.

  $\Delta E=q+w$

The change in entropy is calculated as follows:

  $\Delta S=nR\ln \left(\frac{P_1}{P_2}\right)$

Also, change in Gibbs free energy is related to change in enthalpy and entropy as follows:

  $\Delta G=\Delta H-T.\Delta S$

w − work done

P − pressure

V − volume

n − number of moles

R − universal gas constant

T − temperature

ΔE − energy change

q − heat

ΔS − entropy change

ΔG − Gibbs free energy change

ΔH − enthalpy change

Answer to Problem 147CP

	q	w	ΔE	ΔS	ΔH	ΔG
Step 1	1480 J	-1480 J	0	7.56 J/K	0	-2250 J
Step 2	1240 J	-1240 J	0	5.77 J/K	0	-1720 J
Step 3	1246 J	-1246 J	0	5.81 J/K	0	-1730 J
Total	3966 J	-3966 J	0	19.14 J/K	0	-5700 J

Explanation of Solution

Given information :

1.00 mole of an ideal gas is isothermally expanded at 25 ⁰C as follows.

Step 1: $\text{from }2.45\times {10}^{-2}\text{ atm to 9}\text{.87}\times {\text{10}}^{-3}\text{ atm}$

Step 2: $\text{from 9}\text{.87}\times {\text{10}}^{-3}\text{ atm to 4}\text{.93}\times {\text{10}}^{-3}\text{ atm}$

Step 3: $\text{from 4}\text{.93}\times {\text{10}}^{-3}\text{ atm to 2}\text{.45}\times {\text{10}}^{-3}\text{ atm}$

Step 1:

  $\Delta E=0\text{ and }\Delta H=0\text{ since }\Delta \text{T = 0}$

  $\begin{array}{l}\Delta V=1.00\text{ mol }\times \text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 298 K }\left(\frac{1}{9.87\times {10}^{-3}\text{ atm}}-\frac{1}{2.45\times {10}^{-2}\text{ atm}}\right)\\ \text{ = 1480 L}\end{array}$

  $\begin{array}{l}w=-P.\Delta V\\ w=-9.87\times {10}^{-3}\text{ atm }\times \text{ 1480 L = }-14.6\text{ L}\text{.atm}\\ w=-14.6\text{ L}\text{.atm }\times \text{101}\text{.3 J/L}\text{.atm = }-1480\text{ J}\end{array}$

  $q=-w=1480\text{ J}$

  $\begin{array}{l}\Delta S=1.00\text{ mol }\times \text{ 8}\text{.314 J/mol}\text{.K }\times \text{ ln}\left(\frac{2.45\times {10}^{-2}\text{ atm}}{9.87\times {10}^{-2}\text{ atm}}\right)\\ \text{ = 7}\text{.56 J/K}\end{array}$

  $\begin{array}{l}\Delta G=0-298\text{ K }\times \text{ 7}\text{.56 J/K}\\ \text{ = }-2250\text{ J}\end{array}$

Step 2:

  $\Delta E=0\text{ and }\Delta H=0\text{ since }\Delta \text{T = 0}$

  $\begin{array}{l}\Delta V=1.00\text{ mol }\times \text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 298 K }\left(\frac{1}{4.93\times {10}^{-3}\text{ atm}}-\frac{1}{9.87\times {10}^{-3}\text{ atm}}\right)\\ \text{ = 2483 L}\end{array}$

  $\begin{array}{l}w=-P.\Delta V\\ w=-4.93\times {10}^{-3}\text{ atm }\times \text{ 2483 L = }-12.2\text{ L}\text{.atm}\\ w=-12.2\text{ L}\text{.atm }\times \text{101}\text{.3 J/L}\text{.atm = }-1240\text{ J}\end{array}$

  $q=-w=1240\text{ J}$

  $\begin{array}{l}\Delta S=1.00\text{ mol }\times \text{ 8}\text{.314 J/mol}\text{.K }\times \text{ ln}\left(\frac{9.87\times {10}^{-3}\text{ atm}}{4.93\times {10}^{-3}\text{ atm}}\right)\\ \text{ = 5}\text{.77 J/K}\end{array}$

  $\begin{array}{l}\Delta G=0-298\text{ K }\times \text{ 5}\text{.77 J/K}\\ \text{ = }-1720\text{ J}\end{array}$

Step 3:

  $\Delta E=0\text{ and }\Delta H=0\text{ since }\Delta \text{T = 0}$

  $\begin{array}{l}\Delta V=1.00\text{ mol }\times \text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 298 K }\left(\frac{1}{2.45\times {10}^{-3}\text{ atm}}-\frac{1}{4.93\times {10}^{-3}\text{ atm}}\right)\\ \text{ = 5021 L}\end{array}$

  $\begin{array}{l}w=-P.\Delta V\\ w=-2.45\times {10}^{-3}\text{ atm }\times \text{ 5021 L = }-12.3\text{ L}\text{.atm}\\ w=-12.3\text{ L}\text{.atm }\times \text{101}\text{.3 J/L}\text{.atm = }-1246\text{ J}\end{array}$

  $q=-w=1246\text{ J}$

  $\begin{array}{l}\Delta S=1.00\text{ mol }\times \text{ 8}\text{.314 J/mol}\text{.K }\times \text{ ln}\left(\frac{4.93\times {10}^{-3}\text{ atm}}{2.45\times {10}^{-3}\text{ atm}}\right)\\ \text{ = 5}\text{.81 J/K}\end{array}$

  $\begin{array}{l}\Delta G=0-298\text{ K }\times \text{ 5}\text{.81 J/K}\\ \text{ = }-1730\text{ J}\end{array}$

	q	w	ΔE	ΔS	ΔH	ΔG
Step 1	1480 J	-1480 J	0	7.56 J/K	0	-2250 J
Step 2	1240 J	-1240 J	0	5.77 J/K	0	-1720 J
Step 3	1246 J	-1246 J	0	5.81 J/K	0	-1730 J
Total	3966 J	-3966 J	0	19.14 J/K	0	-5700 J

Conclusion

Thus,

	q	w	ΔE	ΔS	ΔH	ΔG
Step 1	1480 J	-1480 J	0	7.56 J/K	0	-2250 J
Step 2	1240 J	-1240 J	0	5.77 J/K	0	-1720 J
Step 3	1246 J	-1246 J	0	5.81 J/K	0	-1730 J
Total	3966 J	-3966 J	0	19.14 J/K	0	-5700 J