Chapter 10, Problem 155MP

(a)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the expansion is free expansion.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For free expansion process: $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=0\text{ J and }q=0\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 2.00 atm

Initial volume = 2.00 L

Final pressure = 1.00 atm

Final volume = 4.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

Also, since it is given that the gas is undergoing free expansion, thus, there is no external pressure against the gas.

Thus, work is also equal to zero.

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is also equal to zero.

Thus,

  $q=0\text{ J}$

Hence, for free expansion process: $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=0\text{ J and }q=0\text{ J}$

(b)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the expansion is done in one step.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For expansion done in one step: $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-202.6\text{ J and }q=202.6\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 2.00 atm

Initial volume = 2.00 L

Final pressure = 1.00 atm

Final volume = 4.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

Now, work against an external pressure is calculated as:

  $w=-P_{ext}\Delta V$

Put the values,

  $w=-(1.00\text{ atm)}\times (4.00\text{ L}-2.00\text{ L)}\times \left(101.3\frac{\text{J}}{\text{L}\cdot \text{atm}}\right)$

  $=-202.6\text{ J}$

Thus, work done in one step expansion is $-202.6\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $-202.6\text{ J}$ .

Thus,

  $q=202.6\text{ J}$

Therefore, heat change in one step expansion is $202.6\text{ J}$ .

Hence, for expansion done in one step; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-202.6\text{ J and }q=202.6\text{ J}$

(c)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the expansion is done in two steps having volume 3.00 L as the intermediate step.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For expansion done in two steps; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-235.729\text{ J and }q=235.729\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 2.00 atm

Initial volume = 2.00 L

Final pressure = 1.00 atm

Final volume = 4.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

Now, work against an external pressure is calculated as:

  $w=-P_{ext}\Delta V$

Pressure at the end of the first step is calculated as:

  ${\text{P}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}$

Put the values,

  ${\text{P}}_{\text{2}}\text{=}\frac{\text{(2}\text{.00 atm)(2}\text{.00 L)}}{(3.00\text{ L)}}$

  $\text{=}1.33\text{ atm}$

Work done for each step is calculated as:

  $w_1=-\left(1.33\text{ atm}\right)\left(4.00-3.00\text{ L}\right)\left(101.3\frac{\text{J}}{\text{L}\cdot \text{atm}}\right)$

  $=-134.729\text{ J}$

  $w_1=-\left(1.00\text{ atm}\right)\left(3.00-2.00\text{ L}\right)\left(101.3\frac{\text{J}}{\text{L}\cdot \text{atm}}\right)$

  $=-101\text{ J}$

Now, total work done is calculated as:

Total work done = $-134.729\text{ J}-101\text{ J}$

  = $-235.729\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $-235.729\text{ J}$ .

Thus,

  $q=235.729\text{ J}$

Therefore, heat change for expansion intwo steps is $235.729\text{ J}$ .

Hence, for expansion done in two steps; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-235.729\text{ J and }q=235.729\text{ J}$

(d)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the expansion is reversible.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For expansion done in reversible; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-281\text{ J and }q=281\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 2.00 atm

Initial volume = 2.00 L

Final pressure = 1.00 atm

Final volume = 4.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

The mathematical expression for work done for a reversible process (expansion) is:

  $w=-nRT\ln \frac{V_2}{V_1}$

Ideal gas equation is:

  $PV=nRT$

  $w=-\text{PV}\ln \frac{V_2}{V_1}$

Put the values,

  $w=-\text{(1}\text{.00 atm) (4}\text{.00 L) }\left(\text{101}\text{.3 }\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\ln \frac{4.00\text{ L}}{2.00\text{ L}}$

  $=-281\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $-281\text{ J}$ .

Thus,

  $q=281\text{ J}$

Therefore, heat change when expansion is reversible is $235.729\text{ J}$ .

Hence, for expansion is reversible; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=-281\text{ J and }q=281\text{ J}$

(e)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the compression done in one step.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For compression done in one step; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=405\text{ J and }q=-405\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 1.00 atm

Initial volume = 4.00 L

Final pressure = 2.00 atm

Final volume = 2.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

The mathematical expression for work done is:

  $w=-P_{ext}\Delta V$

Put the values,

  $w=-\text{(2}\text{.00 atm) (4}\text{.00 L}-\text{2}\text{.00 L) }\left(\text{101}\text{.3 }\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)$

  $=405\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $405\text{ J}$ .

Thus,

  $q=-405\text{ J}$

Therefore, heat change for compression in one step is $-405\text{ J}$ .

Hence, for compression done in one step; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=405\text{ J and }q=-405\text{ J}$

(f)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the compression done in two steps having volume 3.00 L as intermediate step.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For compression done in two steps; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=338\text{ J and }q=-338\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 1.00 atm

Initial volume = 4.00 L

Final pressure = 2.00 atm

Final volume = 2.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

The mathematical expression for work done is:

  $w=-P_{ext}\Delta V$

The work done for each step is calculated as:

Put the values,

  $w_1=-(1.33\text{ atm)(2}\text{.00 L}-\text{3}\text{.00 L)}\left(\text{101}\text{.3 }\frac{\text{J}}{\text{l}\cdot \text{atm}}\right)$

  $=135\text{ J}$

  $w_1=-(2.00\text{ atm)(3}\text{.00 L}-\text{4}\text{.00 L)}\left(\text{101}\text{.3 }\frac{\text{J}}{\text{l}\cdot \text{atm}}\right)$

  $=203\text{ J}$

Total work done = $203\text{ J + 135 J}$

  = $338\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $338\text{ J}$ .

Thus,

  $q=-338\text{ J}$

Therefore, heat change for compression in two steps is $-338\text{ J}$ .

Hence, for compression in two steps; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=338\text{ J and }q=-338\text{ J}$

(g)

Interpretation Introduction

Interpretation:

The value of w, q, $\text{ΔE},\text{and ΔH}$ should be calculated when the compression is reversible. Also, the answers for both expansion and compression should be compared and implications should be discussed especially considering the changes to the system and to the surroundings which takes place even though the system bring back to initial state.

Concept Introduction:

The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.

The mathematical expression of internal energy is:

  $\Delta E=q+w$

Where, E = internal energy, q = heat and w = work done by or on the system

Thermodynamic work done is equal to $-P\Delta V$ .

Work done on the system is positive whereas the work done by the system is negative.

Answer to Problem 155MP

For compression is reversible; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=281\text{ J and }q=-281\text{ J}$

Explanation of Solution

Given information:

Initial pressure = 1.00 atm

Initial volume = 4.00 L

Final pressure = 2.00 atm

Final volume = 2.00 L

Since, it is given that gas is an ideal gas.

Both Internal energy and enthalpy depends upon temperature.

Therefore,

In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.

The mathematical expression for work done for a reversible process (compression) is:

  $w=nRT\ln \frac{V_2}{V_1}$

Ideal gas equation is:

  $PV=nRT$

  $w=\text{PV}\ln \frac{V_2}{V_1}$

Put the values,

  $w=\text{(1}\text{.00 atm) (4}\text{.00 L) }\left(\text{101}\text{.3 }\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\ln \frac{4.00\text{ L}}{2.00\text{ L}}$

  $=281\text{ J}$

Now, change in internal energy is equal to the sum of work done and heat.

  $\Delta E=q+w$

Since, change in internal energy is equal to zero.

  $0=q+w$

  $q=-w$

Since, work done is equal to $281\text{ J}$ .

Thus,

  $q=-281\text{ J}$

Therefore, heat change when compression is reversible is $-281\text{ J}$ .

Hence, for compression is reversible; $\Delta E=0\text{ J, }\Delta H=0\text{ J},w=281\text{ J and }q=-281\text{ J}$

From above calculations, it is clear that more heat is released for the irreversible compression steps in comparison to irreversible expansion steps.

Since, the system returned to its original state; thus, $\Delta S_{sys}$ as a state function is zero. $\Delta S_{surr}$ ismore than zero because more heat is entered the surroundings through irreversible process. Thus, $\Delta S_{univ}$ is more than zero for two irreversible processes, and the whole process is spontaneous.

Now, for reversible process; during expansion, heat added to the system is equal to the heat lost during compression process. Thus, $\Delta S_{surr}$ is equal to zero because no heat is gain or loss by the surrounding. Since, the system returned to its original state, $\Delta S_{sys}$ as a state function is zero. $\Delta S_{univ}$ isequal to zero for reversible process.