Chapter 3, Problem 3D.1P

(a)

Interpretation Introduction

Interpretation:

The final pressure and temperature of the perfect gas in section A present in a cylinder which is divided into two sections by a frictionless piston has to be calculated.

Concept introduction:

The pressure and temperature of a system are state functions.  The state functions depend only upon the initial and final states of a process and do not depend upon the path taken.

Answer to Problem 3D.1P

The final pressure of the perfect gas in section A present in a cylinder which is divided into two sections by a frictionless piston is $\underset{\_}{49.26atm}$.  The final temperature of the perfect gas in section A present in a cylinder which is divided into two sections by a frictionless piston is $\underset{\_}{900K}$.

Explanation of Solution

The pressure of perfect gas ($p$) is expressed as,

    $p=\frac{nRT}{V}$                                                                                                       (1)

Where

• $n$ is the number of moles of the perfect gas.
• $T$ is the temperature of the perfect gas.
• $V$ is the volume of the perfect gas.
• $R$ is the gas constant $\left(0.0821{\text{dm}}^{\text{3}}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\right)$.

The total volume of the container is $\text{4}{\text{dm}}^{\text{3}}$.

The number of moles of the perfect gas in section B $\left(n_{\text{B}}\right)$ is $\text{2}\text{mol}$.

The temperature of the gas in section B $\left(T_{\text{B}}\right)$ is $\text{300}\text{K}$

The final volume of the perfect gas in section B $\left(V_{\text{B}}\right)$ is $\text{1}\text{.00}{\text{dm}}^{\text{3}}$.

Substitute the value of $n_{\text{B}}$, $T_{\text{B}}$, $V_{\text{B}}$, $R$ in (1).

    $\begin{array}{c}{p}_{\text{B}}=\frac{2\text{mol}\times 0.0821{\text{dm}}^{\text{3}}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times \text{300}\text{K}}{1.00{\text{dm}}^{\text{3}}}\\ =49.26\text{atm}\end{array}$

Since,

    $p_{\text{A}}=p_{\text{B}}$

Pressure in section A is $\underset{\_}{49.26atm}$.

The final temperature in section A can be calculated as,

    $T_{\text{A}}=\frac{p_{\text{A}}{V}_{\text{A}}}{n_{\text{A}}R}$                                                                                                     (2)

The number of moles of the perfect gas in section A $\left(n_{\text{A}}\right)$ is $\text{2}\text{mol}$.

The pressure of the gas in section A is $\left(p_{\text{A}}\right)$ is $\text{49}\text{.26}\text{atm}$

The volume of the perfect gas in section A $\left(V_{\text{A}}\right)$ is $\text{3}\text{.00}{\text{dm}}^{\text{3}}$.

Substitute the value of $n_{\text{A}}$, $p_{\text{A}}$, $V_{\text{A}}$, $R$ in (2).

    $\begin{array}{c}{T}_{\text{B}}=\frac{\text{49}\text{.26}\text{atm}\times \text{3}\text{.00}{\text{dm}}^{\text{3}}}{\text{2}\text{mol}\times 0.0821{\text{dm}}^{\text{3}}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}}\\ =\underset{\_}{900K}\end{array}$

Thus the temperature of section A is $\underset{\_}{900K}$

(b)

Interpretation Introduction

Interpretation:

The change in entropy of the gas in section A has to be calculated.

Concept introduction:

Entropy is the degree of disorder in the system.  As the disorder in the system increases the entropy increases.  It is different for an isothermal and isochoric process.

Answer to Problem 3D.1P

The change in entropy of the gas in section A is $\underset{\_}{50.68J{K}^{-1}}$.

Explanation of Solution

The change in entropy $\left(\Delta S_1\right)$ for constant volume process is given as,

    $\Delta S_1=n_{\text{A}}{C}_{\text{V,m}}\ln \frac{T_2}{T_1}$                                                                                         (3)

The initial temperature in section A $\left(T_1\right)$ is $\text{300}\text{K}$.

The final temperature in section A $\left(T_2\right)$ is $\text{900}\text{K}$.

The constant volume heat capacity of the gas $\left(C_{\text{V,m}}\right)$ is $20\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The number of moles of the gas in section A is $\left(n_{\text{A}}\right)$ is $\text{2}\text{mol}$.

Substitute the value of $n_{\text{A}}$, $T_1$ , $T_2$ and $C_{\text{V,m}}$ in equation (3).

    $\begin{array}{c}\Delta S_1=\text{2}\text{mol}\times 20\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\ln \frac{\text{900}\text{K}}{\text{300}\text{K}}\\ =43.94\text{J}{\text{K}}^{-1}\end{array}$

The change in entropy $\left(\Delta S_2\right)$ for constant temperature process is given as,

    $\Delta S_2=n_{\text{A}}R\ln \frac{V_2}{V_1}$                                                                                         (4)

The initial volume in section A $\left(V_1\right)$ is $\text{2}{\text{dm}}^{\text{3}}$.

The final volume in section A $\left(V_2\right)$ is $3{\text{dm}}^{\text{3}}$.

The gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The number of moles of the gas in section A is $\left(n_{\text{A}}\right)$ is $\text{2}\text{mol}$.

Substitute the value of $n_{\text{A}}$, $V_1$ , $V_2$ and $R$ in equation (4).

    $\begin{array}{c}\Delta S_2=\text{2}\text{mol}\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\ln \frac{3{\text{dm}}^{\text{3}}}{2{\text{dm}}^{\text{3}}}\\ =6.74\text{J}{\text{K}}^{-1}\end{array}$

The total entropy in section A $\left(\Delta S_{\text{A}}\right)$ is the sum of the entropies at constant volume and at constant temperature.

    $\Delta S_{\text{A}}=\Delta S_{\text{1}}+\Delta S_{\text{2}}$

Substitute the value of $\Delta S_1$ and $\Delta S_2$ in the above equation.

    $\begin{array}{c}\Delta S_{\text{A}}=43.94\text{J}{\text{K}}^{-1}+6.74\text{J}{\text{K}}^{-1}\\ =\underset{\_}{50.68J{K}^{-1}}\end{array}$

Thus, the total entropy change in section A is $\underset{\_}{50.68J{K}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The entropy change of the gas in section B has to be calculated.

Concept introduction:

Entropy is the degree of disorder in the system.  As the disorder in the system increases the entropy increases.  It is different for an isothermal and isochoric process.

Answer to Problem 3D.1P

The entropy change of the gas in section B is $\underset{\_}{-11.53J{K}^{-1}}$.

Explanation of Solution

In section B the entire process is isothermal and hence takes place at constant temperature.

The entropy change of gas in section B $\left(\Delta S_{\text{B}}\right)$ is thus given as,

    $\Delta S_{\text{B}}=n_{\text{B}}R\ln \frac{V_2}{V_1}$                                                                                            (4)

The initial volume in section B $\left(V_1\right)$ is $\text{2}{\text{dm}}^{\text{3}}$.

The final volume in section B $\left(V_2\right)$ is $1{\text{dm}}^{\text{3}}$.

The gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The number of moles of the gas in section B is $\left(n_{\text{B}}\right)$ is $\text{2}\text{mol}$.

Substitute the value of $n_{\text{B}}$, $V_1$ , $V_2$ and $R$ in equation (4).

    $\begin{array}{c}\Delta S_2=\text{2}\text{mol}\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\ln \frac{1{\text{dm}}^{\text{3}}}{2{\text{dm}}^{\text{3}}}\\ =-\underset{\_}{11.53J{K}^{-1}}\end{array}$

Thus, the total entropy change in section A is $\underset{\_}{-11.53J{K}^{-1}}$

(d)

Interpretation Introduction

Interpretation:

The change in internal energy for each system has to be calculated.

Concept introduction:

The change in internal energy of a system is directly proportional to the difference in temperature.  For an isothermal process, as the temperature remains constant, change in internal energy becomes zero.

Answer to Problem 3D.1P

The change in internal energy in section A is $\underset{\_}{24000J{K}^{-1}}$.  The change in internal energy in section B is $\underset{\_}{0}$.

Explanation of Solution

The change in internal energy $\left(\Delta U\right)$ is given by the expression,

    $\begin{array}{c}\Delta U=n{C}_{\text{V,m}}\text{d}T\\ =n{C}_{\text{V,m}}\left(T_2-T_1\right)\end{array}$                                                                                     (5)

In section A,

The initial temperature in section A $\left(T_1\right)$ is $\text{300}\text{K}$.

The final temperature in section A $\left(T_2\right)$ is $\text{900}\text{K}$.

The constant volume heat capacity of the gas $\left(C_{\text{V,m}}\right)$ is $20\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The number of moles of the gas in section A is $\left(n_{\text{A}}\right)$ is $\text{2}\text{mol}$.

Substitute the value of $n_{\text{A}}$, $T_1$, $T_2$ and $C_{\text{V,m}}$ in equation (5).

     $\begin{array}{c}\Delta U_{\text{A}}=\text{2}\text{mol}\times 20\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \left(\text{900}\text{K}-3\text{00}\text{K}\right)\\ =\text{2}\text{mol}\times 20\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 6\text{00}\text{K}\\ =\underset{\_}{24000J{K}^{-1}}\end{array}$

Thus the change in internal energy in section A is $\underset{\_}{24000J{K}^{-1}}$.

The change in internal energy is $0$ for an isothermal process.

The process in section B is isothermal.

Thus the change in internal energy in section B is $0$.

(e)

Interpretation Introduction

Interpretation:

The value of $\Delta A$ for section B has to be calculated.  The explanation as to why $\Delta A$ for section A cannot be calculated has to be given.

Concept introduction:

Helmholtz free energy is a thermodynamic quantity. It is represented by the symbol $\Delta A$.  It is the measure of the useful work that can be obtained from a system under closed conditions.

Answer to Problem 3D.1P

The value of $\Delta A$ in section B is $\underset{\_}{3459J}$.  The value of $\Delta A$ in section A cannot be calculated as section A undergoes reversible expansion.

Explanation of Solution

The change in Helmholtz free energy $\left(\Delta A\right)$ is given by the expression,

    $\Delta A=\Delta U-T\Delta S$                                                                                     (6)

Where,

• $\Delta U$ is the change in internal energy.
• $\Delta S$ is the change in entropy.
• $T$ is the temperature.

In section A, the expansion takes place reversibly,

For reversible process, the Helmholtz free energy is $0.$

Hence the Helmholtz free energy cannot be calculated in section A.

In section B,

The change in internal energy ($\Delta U$) is $0$.

The temperature is $\text{300}\text{K}$.

The change in entropy $\left(\Delta S\right)$ is $-\text{11}\text{.53}\text{J}{\text{K}}^{\text{-1}}$.

Substitute the value of $T$, $\Delta U$, and $\Delta S$ in equation (6).

     $\begin{array}{c}\text{ΔA=0}-\text{300}\text{K}\times \left(-\text{11}\text{.53}\text{J}{\text{K}}^{\text{-1}}\right)\\ =\underset{\_}{3459J}\end{array}$

Thus the change in Helmholtz free energy in section B is $\underset{\_}{3459J}$.

(f)

Interpretation Introduction

Interpretation:

The implication of $\Delta A$ for the total process has to be explained.

Concept introduction:

Helmholtz free energy is a thermodynamic quantity. It is represented by the symbol $\Delta A$.  It is the measure of the useful work that can be obtained from a system under closed conditions.

Answer to Problem 3D.1P

The work done in the total process is minimized.

Explanation of Solution

The change in Helmholtz free energy in section B is $\text{3459}\text{J}$.

The change in Helmholtz free energy in section A is $\text{0}$.

Thus the total change in Helmholtz free energy is equal to change in section B.

    $\text{Δ}A=\text{3459}\text{J}$

The change in Helmholtz energy is the maximum amount of work that can be performed.

As the temperature is constant, $\Delta A$ of the reaction is minimized.

Thus the work of the entire system is minimized.