Chapter 10, Problem 136CP

(a)

Interpretation Introduction

Interpretation:

The expansion of gas should be shown as spontaneous when the external pressure is suddenly changes to 2.0 atm.

Concept Introduction:

The mathematical expression for entropy change in system is:

  $\Delta S_{sys}=nR\ln \frac{V_2}{V_1}$

Where, n = number of moles

R = universal gas constant

V₁ and V₂ = initial and final volume

Ideal gas equation is:

  $PV=nRT$

Where, P = pressure

V = volume

n = number of moles

R = Universal gas constant

T = temperature

Answer to Problem 136CP

The process is spontaneous as entropy change of the universe is greater than zero.

Explanation of Solution

Number of moles of monoatomic ideal gas = 1.0 mole

Initial volume = 5.0 L

Initial pressure = 5.0 atm

Final pressure = 2.0 atm

From ideal gas equation,

  $P_1{V}_1=nRT$

Rearrange the above equation in terms of temperature,

  $T=\frac{P_1{V}_1}{nR}$

Put the values,

  $T=\frac{(5.0\text{ atm)(5}\text{.0 L)}}{(1.0\text{ mol)}\left(\text{0}\text{.08206 }\frac{\text{L}\cdot \text{atm}}{\text{K}\cdot \text{mol}}\right)}$

  $T=300\text{ K}$

The final volume is calculated as:

  $V_2=\frac{nRT}{P_2}$

Put the values,

  $V_2=\frac{1.0\text{ mole}\times \text{0}\text{.08206}\frac{\text{L}\cdot \text{atm}}{\text{K}\cdot \text{mole}}\times 300\text{ K}}{2.0\text{ atm}}$

= $12\text{ L}$

The entropy change is calculated as:

  $\Delta S_{sys}=nR\ln \frac{V_2}{V_1}$

Put the values,

  $\Delta S_{sys}=1.0\text{ mole}\times 8.3145\text{ J/K}\cdot \text{mol}\ln \frac{12\text{ L}}{5.0\text{ L}}$

  $\Delta S_{sys}=7.3\text{ J/K}$

Now,

Since, the change in internal energy of an isothermal process is zero, thus heat is equal to negative of work done.

  $q=-w$

  $W=-P_{ext}\Delta V$

  $q=P_{ext}\Delta V$

Put the values,

  $q=2.0\text{ atm}\times \left(12.0\text{ L}-5.0\text{ L}\right)\times \frac{101.3\text{ J}}{\text{L}\cdot \text{atm}}$

  $q=1400\text{ J}$

Entropy change for surrounding is calculated as:

  $\Delta S_{surr}=-\frac{q}{T}$

Put the values,

  $\Delta S_{surr}=-\frac{1400\text{ J}}{300\text{ K}}$

  $\Delta S_{surr}=-4.7\text{ J/K}$

The entropy change of universe is calculated as:

  $\Delta S_{uni}=\Delta S_{sys}+\Delta S_{surr}$

  $\Delta S_{uni}=+7.3\text{ J/K}-4.7\text{ J/K}$

  $\Delta S_{uni}=2.6\text{ J/K}$

Thus, from above value it is clear that the process is spontaneous as entropy change of the universe is greater than zero.

(b)

Interpretation Introduction

Interpretation:

The compression of gas should be shown as spontaneous when the external pressure is suddenly changes back to 5.0 atm.

Concept Introduction:

The mathematical expression for entropy change in system is:

  $\Delta S_{sys}=nR\ln \frac{V_2}{V_1}$

Where, n = number of moles

R = universal gas constant

V₁ and V₂ = initial and final volume

Ideal gas equation is:

  $PV=nRT$

Where, P = pressure

V = volume

n = number of moles

R = Universal gas constant

T = temperature

Answer to Problem 136CP

The process is spontaneous as entropy change of the universe is greater than zero.

Explanation of Solution

Number of moles of monoatomic ideal gas = 1.0 mole

Initial volume = 12.0 L

Final Volume = 12.0 L

From ideal gas equation,

  $P_1{V}_1=nRT$

Rearrange the above equation in terms of temperature,

  $T=\frac{P_1{V}_1}{nR}$

Put the values,

  $T=\frac{(5.0\text{ atm)(5}\text{.0 L)}}{(1.0\text{ mol)}\left(\text{0}\text{.08206 }\frac{\text{L}\cdot \text{atm}}{\text{K}\cdot \text{mol}}\right)}$

  $T=300\text{ K}$

The initial volume is calculated as:

  $V_1=\frac{nRT}{P_1}$

Put the values,

  $V_1=\frac{1.0\text{ mole}\times \text{0}\text{.08206}\frac{\text{L}\cdot \text{atm}}{\text{K}\cdot \text{mole}}\times 300\text{ K}}{2.0\text{ atm}}$

= $12\text{ L}$

The entropy change is calculated as:

  $\Delta S_{sys}=nR\ln \frac{V_2}{V_1}$

Put the values,

  $\Delta S_{sys}=1.0\text{ mole}\times 8.3145\text{ J/K}\cdot \text{mol}\ln \frac{5.0\text{ L}}{12.0\text{ L}}$

  $\Delta S_{sys}=-7.3\text{ J/K}$

Now,

Since, the change in internal energy of an isothermal process is zero, thus heat is equal to negative of work done.

  $q=-w$

  $W=-P_{ext}\Delta V$

  $q=P_{ext}\Delta V$

Put the values,

  $q=5.0\text{ atm}\times \left(5.0\text{ L}-12.0\text{ L}\right)\times \frac{101.3\text{ J}}{\text{L}\cdot \text{atm}}$

  $q=-3500\text{ J}$

Entropy change for surrounding is calculated as:

  $\Delta S_{surr}=-\frac{q}{T}$

Put the values,

  $\Delta S_{surr}=+\frac{3500\text{ J}}{300\text{ K}}$

  $\Delta S_{surr}=12\text{ J/K}$

The entropy change of universe is calculated as:

  $\Delta S_{uni}=\Delta S_{sys}+\Delta S_{surr}$

  $\Delta S_{uni}=-7.3\text{ J/K}+12\text{ J/K}$

  $\Delta S_{uni}=5\text{ J/K}$

Thus, from above value it is clear that the process is spontaneous as entropy change of the universe is greater than zero.

(c)

Interpretation Introduction

Interpretation:

The value of should be calculated along with its sign comparison for part (a) and (b) and also, the reason should be discussed for not using this sign to predict spontaneity.

Concept Introduction:

The mathematical expression for Gibbs free energy change is:

  $\Delta G=\Delta H-T\Delta S$

Where, $\Delta H$ = change in enthalpy

T = temperature

  $\Delta S$ = change in entropy

Explanation of Solution

For isothermal process, change in enthalpy is equal to zero.

Since, both of the process is isothermal, thus the enthalpy change is equal to zero,

Thus, expression of Gibbs free energy is shown as:

  $\Delta G=-T\Delta S$

For part (a) that is expansion process:

Put the values,

  $\Delta G=-300\text{ K}\times 7.3\text{ J/K}$

  $\Delta G=-2190\text{ J}$

In kJ,

  $\Delta G=-2190\text{ J}\times \frac{1\text{ kJ}}{1000\text{ J}}=-2.2\text{ kJ}$

For part (b) that is compression process:

Put the values,

  $\Delta G=-300\text{ K}\times -7.3\text{ J/K}$

  $\Delta G=2190\text{ J}$

In kJ,

  $\Delta G=+2190\text{ J}\times \frac{1\text{ kJ}}{1000\text{ J}}=+2.2\text{ kJ}$

Now, according to the value of changes in entropy in both parts shows that the process is spontaneous as entropy change is greater than zero but the sign of change in Gibbs free energy is different for both parts. This is because change in Gibbs free energy depends on the sign of entropy change of the system only. Therefore, the sign of change in Gibbs free energy cannot be used for prediction of spontaneity.