Chapter 10, Problem 148CP

(a)

Interpretation Introduction

Interpretation:

The value of $q,w,\Delta E,\Delta S,\Delta H\text{ and }\Delta G$ for the expansion of 1.00 mole of an ideal gas at 25 ⁰C isothermally and irreversibly from $2.45\times {10}^{-2}\text{ atm to 2}\text{.45 }\times {\text{10}}^{-3}\text{ atm}$ should be calculated in one step.

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 148CP

  $\begin{array}{l}\Delta E=0\\ \Delta H=0\end{array}$

  $w=-2229\text{ J}$

  $q=2229\text{ J}$

  $\Delta S\text{ = 19}\text{.1 J/K}$

  $\Delta G\text{ = }-5690\text{ J}$

Explanation of Solution

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

  $\begin{array}{l}\Delta V=1.00\text{ mol }\times \text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 298 K }\left(\frac{1}{2.45\times {10}^{-3}\text{ atm}}-\frac{1}{2.45\times {10}^{-2}\text{ atm}}\right)\\ \text{ = 8983 L}\end{array}$

  $\begin{array}{l}w=-P.\Delta V\\ w=-2.45\times {10}^{-3}\text{ atm }\times \text{ 8983 L = }-22.0\text{ L}\text{.atm}\\ w=-22.0\text{ L}\text{.atm }\times \text{101}\text{.3 J/L}\text{.atm = }-2229\text{ J}\end{array}$

  $q=-w=2229\text{ J}$

  $\begin{array}{l}\Delta S=1.00\text{ mol }\times \text{ 8}\text{.314 J/mol}\text{.K }\times \text{ ln}\left(\frac{2.45\times {10}^{-2}\text{ atm}}{2.45\times {10}^{-3}\text{ atm}}\right)\\ \text{ = 19}\text{.1 J/K}\end{array}$

  $\begin{array}{l}\Delta G=0-298\text{ K }\times \text{ 19}\text{.1 J/K}\\ \text{ = }-5690\text{ J}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $q,w,\Delta E,\Delta S,\Delta H\text{ and }\Delta G$ for the expansion of 1.00 mole of an ideal gas at 25 ⁰C isothermally and reversibly from $2.45\times {10}^{-2}\text{ atm to 2}\text{.45 }\times {\text{10}}^{-3}\text{ atm}$ should be calculated in one step.

Concept Introduction :

The change in entropy is calculated as follows:

  $\Delta S=\frac{q_{rev}}{T}$

Here,

ΔS − entropy change

q_rev − heat absorbed or release during reversible process

T − temperature

Answer to Problem 148CP

  $\begin{array}{l}\Delta E=0\\ \Delta H=0\end{array}$

  $\Delta S\text{ = 19}\text{.1 J/K}$

  $\Delta G\text{ = }-5690\text{ J}$

  $q_{rev}=5690\text{ J}$

  $w_{rev}=-5690\text{ J}$

Explanation of Solution

  $\begin{array}{l}\Delta E=0\\ \Delta H=0\end{array}$

  $\Delta S\text{ = 19}\text{.1 J/K}$

  $\Delta G\text{ = }-5690\text{ J}$ same as in part a, because these are state functions.

  $\begin{array}{l}\Delta S=\frac{q_{rev}}{T}\\ q_{rev}=T.\Delta S=298\text{ K }\times \text{ 19}\text{.1 J/K}\\ q_{rev}=5690\text{ J}\end{array}$

  $\begin{array}{l}\Delta E=0=q+w\\ w_{rev}=-q_{rev}=-5690\text{ J}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $q,w,\Delta E,\Delta S,\Delta H\text{ and }\Delta G$ for the compression of 1.00 mole of an ideal gas at 25 ⁰C isothermally and irreversibly from $\text{2}\text{.45 }\times {\text{10}}^{-3}\text{ atm to }2.45\times {10}^{-2}\text{ atm}$ should be calculated in one step.

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 148CP

  $\begin{array}{l}\Delta E=0\\ \Delta H=0\end{array}$

  $\Delta S\text{ = }-\text{19}\text{.1 J/K}$

  $\Delta G\text{ = }5690\text{ J}$

  $w=22300\text{ J}$

  $q=-22300\text{ J}$

Explanation of Solution

  $\begin{array}{l}\Delta E=0\\ \Delta H=0\end{array}$

  $\Delta S\text{ = }-\text{19}\text{.1 J/K}$

  $\Delta G\text{ = }5690\text{ J}$

The signs are opposite as this is the reverse process described in part a.

  $\begin{array}{l}w=-2.45\times {10}^{-2}\text{ atm }\times 1.00\text{ mol }\times \text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\left(\frac{1}{2.45\times {10}^{-2}\text{ atm}}-\frac{1}{2.45\times {10}^{-3}\text{ atm}}\right)\times 101.3\text{ J/L}\text{.atm}\\ w=22300\text{ J}\end{array}$

  $\begin{array}{l}q=-w\\ q=-22300\text{ J}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The PV diagrams for the process described in parts a, b, and c should be constructed.

Concept Introduction:

A PV diagram, a pressure-volume diagram is used to describe corresponding changes in pressure and volume of a system.

  

Answer to Problem 148CP

Explanation of Solution

Part a − isothermal irreversible expansion

Part b − isothermal reversible expansion

Part c − isothermal irreversible compression

  

(e)

Interpretation Introduction

Interpretation:

The entropy change in the surroundings for the process described in parts a,b and c should be calculated.

Concept Introduction :

  $\Delta S_{surr}=-\frac{q_{actual}}{T}$

ΔS − entropy change

q − heat absorbed or released

T − temperature

Answer to Problem 148CP

  $\Delta S_{surr,a}=-7.48\text{ J/K}$

  $\Delta S_{surr,b}=-19.1\text{ J/K}$

  $\Delta S_{surr,c}=74.8\text{ J/K}$

Explanation of Solution

The change in entropy for each step is calculated as follows:

  $\Delta S_{surr,a}=-\frac{2229\text{ J}}{298\text{ K}}=-7.48\text{ J/K}$

  $\Delta S_{surr,b}=-\Delta S=-19.1\text{ J/K}$

  $\Delta S_{surr,c}=\frac{22300\text{ J}}{298\text{ K}}=74.8\text{ J/K}$