Chapter 3, Problem 3A.1P

(a)

Interpretation Introduction

Interpretation: The value of $q$, $w$, $\Delta U$, $\Delta H$, $\Delta S$, $\Delta S_{\text{surr}}$, $\Delta S_{\text{tot}}$ for reversible isothermal expansion of a sample of 1.00 mol of perfect gas molecules at $27°\text{C}$ from an initial pressure of $3.00\text{atm}$ to a final pressure of $1.00\text{atm}$ has to be calculated.

Concept introduction: An isothermal expansion is carried out in such a way that the temperature remains constant throughout the process while the volume expands from an initial value to a final value and pressure reduces from an initial value to a final value.  For a reversible process, the value of $p_{\text{ext}}=p_{\text{int}}$ and can be integrated.

Answer to Problem 3A.1P

For reversible isothermal expansion of the perfect gas, the value of $q$ is $\underset{\_}{27.4\times 10^{3}J}$, the value of $w$ is $\underset{\_}{-27.4\times 10^{3}J}$, the value of $\Delta U$ is $\underset{\_}{0}$, the value of $\Delta H$ is $\underset{\_}{0}$, the value of $\Delta S_{\text{sys}}$ is $\underset{\_}{9.13J{K}^{-1}}$, the value of $\Delta S_{\text{surr}}$ is $\underset{\_}{-9.13J{K}^{-1}}$, the value of $\Delta S_{\text{total}}$  is $\underset{\_}{0}$

Explanation of Solution

According to the first law of thermodynamics,

    $\Delta U=q+w$

For a reversible isothermal process, the temperature remains constant.

Hence, the value of $\Delta U=0$.  Thus,

  $q=-w$                                                                                                           (1)

The work done $\left(w\right)$ in a reversible isothermal expansion is given by the relation,

    $w=-2.303nRT\log \frac{p_{\text{i}}}{p_{\text{f}}}$                                                                                  (2)

Where

• • $n$ is the number of moles of the prefect gas
• • $R$ is the gas constant $\left(8.31{\text{4 J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• • $T$ is the temperature of the perfect gas.
• • $p_{\text{i}}$ is the initial pressure.
• • $p_{\text{f}}$ is the final pressure.

The number of moles $\left(n\right)$ is $1.0\text{mol}$.

The temperature is $27°\text{C}$.

The temperature values in degree Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature value in the above equation as follows:

    $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =27+273.15\\ =300.15\text{K}\end{array}$

Initial pressure $\left(p_{\text{i}}\right)$ is $3.00\text{atm}$.

Final pressure $\left(p_{\text{f}}\right)$ is $1.00\text{atm}$.

Substitute the values of $n$, $T$, $p_{\text{i}}$, $p_{\text{f}}$ in equation (2).

    $\begin{array}{c}w=-\text{2}\text{.303×1}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\text{×}\text{300}\text{.15}\text{K}\times \log \frac{3.0\text{atm}}{1.0\text{atm}}\\ =-\underset{\_}{2.74\times 10^{3}J}\end{array}$

Substitute the value of $w$ from equation (1).

    $\begin{array}{c}q=-w\\ =\underset{\_}{2.74\times 10^{3}J}\end{array}$

In a reversible isothermal expansion, temperature remains constant.  Hence,

  $\Delta H=\underset{\_}{0}$.

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation,

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$                                                                                     (3)

Where,

• • $\Delta S_{\text{total}}$ is the total change in entropy.
• • $\Delta S_{\text{sys}}$ is the change in entropy for system.
• • $\Delta S_{\text{surr}}$ is the change in entropy for surroundings.

For a reversible isothermal expansion, the total entropy change is zero. Therefore

  $\Delta S_{\text{total}}=\underset{\_}{0}$

Substitute the value of $\Delta S_{\text{total}}$ in equation (1).  Substitute the value of $\Delta S_{\text{total}}$ in equation (3).

  $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ 0=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\end{array}$

Hence,

  $\Delta S_{\text{surr}}=-\Delta S_{\text{sys}}$                                                                                               (4)

The expression for entropy change from equation (2) is given as,

    $\Delta S_{\text{sys}}=2.303nR\log \frac{p_{\text{i}}}{p_{\text{f}}}$

Where

• • $n$ is the number of moles of the prefect gas
• • $R$ is the gas constant $\left(8.31{\text{4 J K}}^{-1}{\text{ mol}}^{-1}\right)$
• • $p_{\text{i}}$ is the initial pressure
• • $p_{\text{f}}$ is the final pressure

Substitute the value of $n$, $R$, $p_{\text{i}}$, $p_{\text{f}}$ in equation (2).

    $\begin{array}{c}\Delta S_{\text{sys}}=2.303\times 1\text{mol}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times \log \left(\frac{\text{3}\text{.0}\text{atm}}{\text{1}\text{.0}\text{atm}}\right)\\ =\underset{\_}{9.13J{K}^{-1}}\end{array}$

Substitute the value of $\Delta {\text{S}}_{\text{sys}}$ in equation (4).

  $\begin{array}{c}\Delta S_{\text{surr}}=-\Delta S_{\text{sys}}\\ =-\underset{\_}{9.13J{K}^{-1}}\end{array}$

Hence for an isothermal reversible expansion of a perfect gas, the value of $q$ is $27.4\times 10^{3}J$, the value of $w$ is $-27.4\times 10^{3}J$, the value of $\Delta U$ is $\underset{\_}{0}$, the value of $\Delta H$ is $\underset{\_}{0}$, the value of $\Delta {\text{S}}_{\text{sys}}$ is $9.13J{K}^{-1}$, the value of $\Delta S_{\text{surr}}$ is $-9.13J{K}^{-1}$, the value of $\Delta S_{\text{total}}$ is $\underset{\_}{0}$.

(b)

Interpretation Introduction

Interpretation: The value of $q$, $w$, $\Delta U$, $\Delta H$, $\Delta S$, $\Delta S_{\text{surr}}$, $\Delta S_{\text{tot}}$ for  isothermal expansion of a sample of 1.00 mol of perfect gas at $27°\text{C}$ from an initial pressure of $3.00\text{atm}$ to a final pressure of $1.00\text{atm}$ against an external pressure of $1\text{atm}$ has to be calculated.

Concept introduction: An isothermal expansion is carried out in such a way that the temperature remains constant throughout the process while the volume expands from an initial value to a final value and pressure reduces from an initial value to a final value.. In an irreversible process the value of $p_{\text{ext}}\ne p_{\text{int}}$ and cannot be integrated.

Answer to Problem 3A.1P

for isothermal expansion of a perfect gas against constant external pressure of $\text{1}\text{atm}$, the value of $q$ is $\underset{\_}{1.66kJ}$, the value of $w$ is $\underset{\_}{-1.66kJ}$, the value of $\Delta U$ is $\underset{\_}{0}$, the value of $\Delta H$ is $\underset{\_}{0}$, the value of $\Delta S_{\text{sys}}$ is $\underset{\_}{9.13J{K}^{-1}}$, the value of $\Delta S_{\text{surr}}$ is $\underset{\_}{-5.53J{K}^{-1}}$, the value of $\Delta S_{\text{total}}$ is $\underset{\_}{3.6J{K}^{-1}}$.

Explanation of Solution

According to the first law of thermodynamics,

    $\Delta U=q+w$

For an isothermal expansion process, the temperature remains constant.

Hence the value of $\Delta U=0$.  Thus,

  $q=-w$                                                                                                  (1)

The work done $\left(w\right)$ for an isothermal expansion against constant pressure is given by the relation,

    $\begin{array}{c}w=-p_{\text{ext}}\Delta V\\ =-p_{\text{ext}}\left(V_{\text{f}}-V_{\text{i}}\right)\end{array}$                                                                                         (4)

According to the ideal gas equation,

    $V=\frac{nRT}{p}$

Hence,

  $V_{\text{f}}=\frac{nRT}{p_{\text{f}}}$ and $V_{\text{i}}=\frac{nRT}{p_{\text{i}}}$

Substitute the value of $V_{\text{f}}$ and $V_{\text{i}}$ in equation (4).

    $w=-p_{\text{ext}}\left(\frac{nRT}{p_{\text{f}}}-\frac{nRT}{p_{\text{i}}}\right)$                                                                               (5)

Where

• • $p_{\text{ext}}$ is the external pressure.
• • $n$ is the number of moles of the perfect gas.
• • $R$ is the gas constant $\left(\text{0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)$.
• • $T$ is the temperature of the perfect gas.
• • $p_{\text{i}}$ is the initial pressure.
• • $p_{\text{f}}$ is the final pressure.

The number of moles $\left(n\right)$ is $1.0\text{mol}$.

The temperature given $t\left(\text{°C}\right)$ is $27°\text{C}$.

The temperature values in degree Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature value in the above equation as follows.

    $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =27+273.15\\ =300.15\text{K}\end{array}$

Initial pressure $\left(p_{\text{i}}\right)$ is $3.00\text{atm}$.

Final pressure $\left(p_{\text{f}}\right)$ is $1.00\text{atm}$.

Substitute the values of $n$, $T$, $p_{\text{i}}$, $p_{\text{f}}$, $p_{\text{ext}}$, $R$ in equation (5).

    $\begin{array}{c}w=-1.00\text{atm}\left(\begin{array}{l}\left(\frac{\text{1}\text{mol×0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×300}\text{.15}\text{K}}{\text{1}\text{.00}\text{atm}}\right)-\\ \left(\frac{\text{1}\text{mol×0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×300}\text{.15}\text{K}}{\text{3}\text{.00}\text{atm}}\right)\end{array}\right)\\ \text{=}-\text{16}\text{.42}\text{L}\text{atm}\text{×}\frac{\text{101}\text{.33}\text{J}}{\text{1}\text{L}\text{atm}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\\ \text{=}\underset{\_}{-1.66kJ}\end{array}$

Substitute the value of $w$ from equation (1).

    $\begin{array}{c}q=-w\\ =\underset{\_}{1.66kJ}\end{array}$

In a reversible isothermal expansion, temperature remains constant.

Hence, $\Delta H=\underset{\_}{0}$.

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation,

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$                                                                                     (3)

Where,

• • $\Delta S_{\text{total}}$ is the total change in entropy.
• • $\Delta S_{\text{sys}}$ is the change in entropy for system.
• • $\Delta S_{\text{surr}}$ is the change in entropy for surroundings.

The expression for entropy change is given as,

    $\Delta S_{\text{sys}}=2.303nR\log \frac{p_{\text{i}}}{p_{\text{f}}}$                                                                                 (2)

Where

• • $n$ is the number of moles of the prefect gas.
• • $R$ is the gas constant $\left(8.31{\text{4 J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• • $p_{\text{i}}$ is the initial pressure.
• • $p_{\text{f}}$ is the final pressure.

Substitute the value of $n$, $R$, $p_{\text{i}}$, $p_{\text{f}}$ in equation (5).

    $\begin{array}{c}\Delta {\text{S}}_{\text{sys}}=2.303\times 1\text{mol}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times \log \left(\frac{\text{3}\text{.0}\text{atm}}{\text{1}\text{.0}\text{atm}}\right)\\ =\underset{\_}{9.13J{K}^{-1}}\end{array}$

The entropy change for the surroundings $\Delta S_{\text{surr}}$ is given by the equation,

    $\begin{array}{c}\Delta S_{\text{surr}}=-\frac{q}{T}\\ =-\frac{\text{1}\text{.66×}{\text{10}}^{\text{3}}\text{J}}{\text{300}\text{.15}\text{K}}\\ \underset{\_}{=-5.53J{K}^{-1}}\end{array}$

The total entropy change $\Delta S_{\text{total}}$ is given by the equation (3)

    $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ =\text{9}\text{.13}{\text{JK}}^{\text{-1}}\text{+}\left(-\text{5}\text{.53}{\text{JK}}^{\text{-1}}\right)\\ ={\underset{\_}{3.6JK}}^{\text{-1}}\end{array}$

Hence, for isothermal expansion of a perfect gas against constant external pressure of $\text{1}\text{atm}$, the value of $q$ is $\underset{\_}{1.66kJ}$, the value of $w$ is $\underset{\_}{-1.66kJ}$, the value of $\Delta U$ is $\underset{\_}{0}$, the value of $\Delta H$ is $\underset{\_}{0}$, the value of $\Delta S_{\text{sys}}$ is $\underset{\_}{9.13J{K}^{-1}}$, the value of $\Delta S_{\text{surr}}$ is $\underset{\_}{-5.53J{K}^{-1}}$, the value of $\Delta S_{\text{total}}$ is $\underset{\_}{3.6J{K}^{-1}}$.