Chapter 15, Problem 91P

(a)

To determine

The heat flow in or out of the gas during each step.

Answer to Problem 91P

The heat flow in or out of the gas during each step are $\underset{\_}{36.5\text{kJ}}$, $\underset{\_}{-36.3\text{kJ}}$ and $\underset{\_}{21.8\text{kJ}}$.

Explanation of Solution

Write the expression for the heat for the isothermal expansion.

$Q_1=\Delta U_1-W_{\text{input}}$ (I)

Here, $Q_1$ is the heat for the first process of isothermal expansion of the gas, $\Delta U_1$ is the change in the internal energy for the process and $W_{\text{input}}$ is the input work done.

Write the expression for the heat for the isobaric compression.

$Q_2=\Delta U_2-W_{\text{input}}$ (II)

Here, $Q_2$ is the heat for the second process of isobaric compression of the gas, $\Delta U_2$ is the change in the internal energy for the process and $W_{\text{input}}$ is the input work done.

Write the expression for the heat for the isochoric process.

$Q_3=\Delta U_3-W_{\text{input}}$ (III)

Here, $Q_3$ is the heat for the third isochoric process of the gas, $\Delta U_3$ is the change in the internal energy for the process and $W_{\text{input}}$ is the input work done.

Write the expression for contraction temperature.

$T_{\text{2f}}=T_{2\text{i}}\left(\frac{V_{\text{2f}}}{V_{\text{2i}}}\right)$ (IV)

Here, $T_{\text{2f}}$ is the final temperature, $T_{\text{2i}}$ is the initial temperature, $V_{\text{2f}}$ is the final volume, $V_{\text{2i}}$ is the initial volume.

Conclusion:

Substitute $0$ for $\Delta U_1$ and $-36.5\text{kJ}$ for $W_{\text{input}}$ in equation (I).

$\begin{array}{c}{Q}_1=0-\left(-36.5\text{kJ}\right)\\ =36.5\text{kJ}\end{array}$

Substitute $650.0\text{K}$ for $T_{\text{2i}}$ and $\frac{1}{9.50}$ for $\frac{V_{\text{2f}}}{V_{\text{2i}}}$ in equation (IV).

$\begin{array}{c}{T}_{\text{2f}}=650.0\text{K}\left(\frac{1}{9.50}\right)\\ =68.4\text{K}\end{array}$

Substitute $\frac{3}{2}nR\Delta T$ for $\Delta U_2$ and $-nR\Delta T$ for $W_{\text{input}}$ in equation (II).

$\begin{array}{c}{Q}_2=\frac{3}{2}nR\Delta T-\left(-nR\Delta T\right)\\ =\frac{5}{2}nR\Delta T\end{array}$

Here, $n$ is the number of moles, $R$ is the universal gas constant and $\Delta T$ is the change in temperature.

Substitute $3.00\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $68.4\text{K}-650\text{K}$ for $\Delta T$ in the above equation.

$\begin{array}{c}{Q}_2=\frac{5}{2}\left(3.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(68.4\text{K}-650\text{K}\right)\\ =-36.3\text{kJ}\end{array}$

Substitute $\frac{3}{2}nR\Delta T$ for $\Delta U_3$ and $0$ for $W_{\text{input}}$ in equation (III).

$\begin{array}{c}{Q}_3=\frac{3}{2}nR\Delta T-0\\ =\frac{3}{2}nR\Delta T\end{array}$

Substitute $3.00\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $650\text{K}-68.4\text{K}$ for $\Delta T$ in the above equation.

$\begin{array}{c}{Q}_3=\frac{3}{2}\left(3.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(650\text{K}-68.4\text{K}\right)\\ =21.8\text{kJ}\end{array}$

Therefore, the heat flow in or out of the gas during each step are $\underset{\_}{36.5\text{kJ}}$, $\underset{\_}{-36.3\text{kJ}}$ and $\underset{\_}{21.8\text{kJ}}$.

(b)

To determine

The entropy change of the gas during the isothermal step.

Answer to Problem 91P

The entropy change of the gas during the isothermal step is $\underset{\_}{+56.2\text{J}/\text{K}}$.

Explanation of Solution

Write the expression to find the change in entropy.

$\Delta S=\frac{Q_1}{T}$

Here, $Q_1$ is the heat in the isothermal step and $T$ is the temperature.

Conclusion:

Substitute $36.5\text{kJ}$ for $Q_1$ and $650.0\text{K}$ for $T$ in the above equation.

$\begin{array}{c}\Delta S=\frac{36.5\text{kJ}}{650.0\text{K}}\\ =+56.2\text{J}/\text{K}\end{array}$

Therefore, the entropy change of the gas during the isothermal step is $\underset{\_}{+56.2\text{J}/\text{K}}$.

(c)

To determine

The entropy change of the gas for the complete cycle.

Answer to Problem 91P

The entropy change of the gas for the complete cycle is $\underset{\_}{\Delta S_{\text{gas}}=0}$.

Explanation of Solution

The entropy is a state variable, it depends on the initial and final values of the variable only. Since the process completes a cycle, the change in entropy of the gas is zero. But to validate the second law of thermodynamics, that the entropy of a system always increases, the entropy of the environment increases, so the system as a whole has increase in entropy.

Therefore, the entropy change of the gas for the complete cycle is $\underset{\_}{\Delta S_{\text{gas}}=0}$.