Chapter 15, Problem 88P

(a)

To determine

The heat flow into or out of the gas through the four steps of the cycle.

Answer to Problem 88P

The heat flow to the system during process $A$ is $692\text{J}$. The heat change of the system during the process $B$ is $-2080\text{J}$. The heat change of the system during the process $C$ is $-506\text{J}$. The heat change during the process $D$ is $2080\text{J}$.

Explanation of Solution

The number of moles of the diatomic gas is $1000\text{mol}$.

The figure below shows the cycle.

During the process $A$, the gas is in contact with heat reservoir at temperature $373\text{K}$. Thus it is a isothermal process. In isothermal process the temperature change is zero, thus the internal energy change is zero. Under this situation the heat flow to the cold reservoir is equal to the work done.

Write the formula for the heat flow out of the system in process $A$.

$Q_A=nR{T}_A\ln \frac{V_f}{V_i}$ (I)

Here, $Q_A$ is the heat flow out of the system in process $A$, $n$ is the number of moles, $R$ is the universal gas constant, $T_A$ is the temperature during process $A$, $V_f$ is the final volume, $V_i$ is the initial volume.

During the constant volume process $B$ heat flows into the cold reservoir. During the constant volume process the work done is zero. Thus the heat change is equal to the change in internal energy of the system.

Write the formula for the heat change of the system during the process $B$.

$Q_B=n{C}_V\Delta T$ (II)

Here, $Q_B$ is the heat change during the process $B$, $C_V$ is the specific heat capacity under constant volume, $\Delta T$ is the change in temperature.

Write the formula for the specific heat capacity under constant volume for diatomic gases.

$C_V=\frac{5}{2}R$

Substitute the above equation in equation (II).

$Q_B=\frac{5}{2}nR\Delta T$ (III)

During the process $C$, the gas is in contact with heat reservoir at temperature $273\text{K}$. Thus it is an isothermal process. In isothermal process the temperature change is zero, thus the internal energy change is zero. Under this situation the heat flow to the cold reservoir is equal to the work done.

Write the formula for the heat flow out of the system in process $C$.

$Q_C=nR{T}_C\ln \frac{V_f}{V_i}$ (IV)

Here, $Q_C$ is the heat flow out of the system in process $C$, $T_C$ is the temperature during process $C$.

During the constant volume process $D$ heat flows out of the hot reservoir.

Refer equation (III) and write the formula for the change in heat for the process $D$.

$Q_D=\frac{5}{2}nR\Delta T$ (V)

Here, $Q_D$ is the change in heat during process $D$.

Conclusion:

Substitute $1000\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $373\text{K}$ for $T_A$, $2.50{\text{m}}^{\text{3}}$ for $V_f$, $2.00{\text{m}}^{\text{3}}$ for $V_i$ in equation (I).

$\begin{array}{c}{Q}_A=\left(1000\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(373\text{K}\right)\ln \frac{2.50{\text{m}}^{\text{3}}}{2.00{\text{m}}^{\text{3}}}\\ =692\text{J}\end{array}$

Substitute $1000\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $273\text{K}-373\text{K}$ for $\Delta T$ in equation (III).

$\begin{array}{c}{Q}_B=\frac{5}{2}\left(1000\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(273\text{K}-373\text{K}\right)\\ =-2080\text{J}\end{array}$

Substitute $1000\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $273\text{K}$ for $T_C$, $2.00{\text{m}}^{\text{3}}$ for $V_f$, $2.50{\text{m}}^{\text{3}}$ for $V_i$ in equation (IV).

$\begin{array}{c}{Q}_C=\left(1000\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(273\text{K}\right)\ln \frac{2.00{\text{m}}^{\text{3}}}{2.50{\text{m}}^{\text{3}}}\\ =-506\text{J}\end{array}$

Substitute $1000\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $373\text{K}-273\text{K}$ for $\Delta T$ in equation (V).

$\begin{array}{c}{Q}_D=\frac{5}{2}\left(1000\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(373\text{K}-273\text{K}\right)\\ =2080\text{J}\end{array}$

The heat flow to the system during process $A$ is $692\text{J}$. The heat change of the system during the process $B$ is $-2080\text{J}$. The heat change of the system during the process $C$ is $-506\text{J}$. The heat change during the process $D$ is $2080\text{J}$.

(b)

To determine

The neat heat change during one cycle.

Answer to Problem 88P

The net heat change in one cycle is $\text{186}\text{J}$.

Explanation of Solution

The number of moles of the diatomic gas is $1000\text{mol}$.

Write the formula for the net heat change during one cycle.

$Q=Q_A+Q_B+Q_C+Q_D$ (VI)

Here, $Q$ is the net heat change in the cycle, $Q_A$ is the heat flow out of the system in process $A$, $Q_B$ is the heat change during the process $B$, $Q_C$ is the heat flow out of the system in process $C$, $Q_D$ is the change in heat during process $D$.

Conclusion:

Substitute $692\text{J}$ for $Q_A$, $-2080\text{J}$ for $Q_B$, $-506\text{J}$ for $Q_C$, $2080\text{J}$ for $Q_D$.

$\begin{array}{c}Q=692\text{J}-2080\text{J}-506\text{J}+2080\text{J}\\ \text{=186}\text{J}\end{array}$

The net heat change in one cycle is $\text{186}\text{J}$.

(C)

To determine

The entropy change of the reservoirs during each of the processes.

Answer to Problem 88P

The change in entropy during process $B$ is $7.61\text{J/K}$. The change in entropy during the process $C$ is $1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $A$ is $-1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $D$ is $-5.57\text{J/K}$.

Explanation of Solution

The number of moles of the diatomic gas is $1000\text{mol}$.

The figure below shows the cycle.

Write the formula for the change in entropy of a system.

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

Here, $\Delta S$ is the change in entropy, $Q$ is the heat flow into the system, $T$ is the temperature.

The change in entropy of the reservoir will be negative of the change in entropy of the syste.

$\Delta S_R=\frac{-Q}{T}$

Here, $\Delta S_R$ is the change in entropy of the reservoir.

Conclusion:

Substitute $-2080\text{J}$ for $Q$, $273\text{K}$ for $T$ to determine the entropy change of the cold reservoir during process $B$.

$\begin{array}{c}\Delta S_R=\frac{-\left(-2080\text{J}\right)}{273\text{K}}\\ =7.61\text{J/K}\end{array}$

Substitute $-506\text{J}$ for $Q$, $273\text{K}$ for $T$ to determine the entropy change of the cold reservoir during process $C$.

$\begin{array}{c}\Delta S_R=\frac{-\left(-506\text{J}\right)}{273\text{K}}\\ =1.86\text{J/K}\end{array}$

Substitute $692\text{J}$ for $Q$, $373\text{K}$ for $T$ to determine the entropy change of the cold reservoir during process $A$.

$\begin{array}{c}\Delta S_R=\frac{-\left(692\text{J}\right)}{373\text{K}}\\ =-1.86\text{J/K}\end{array}$

Substitute $2080\text{J}$ for $Q$, $373\text{K}$ for $T$ to determine the entropy change of the cold reservoir during process $D$.

$\begin{array}{c}\Delta S_R=\frac{-\left(2080\text{J}\right)}{373\text{K}}\\ =-5.57\text{J/K}\end{array}$

The change in entropy during process $B$ is $7.61\text{J/K}$. The change in entropy during the process $C$ is $1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $A$ is $-1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $D$ is $-5.57\text{J/K}$.

(d)

To determine

The net entropy change during one cycle of the universe.

Answer to Problem 88P

The change in entropy during process $B$ is $7.61\text{J/K}$. The change in entropy during the process $C$ is $1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $A$ is $-1.86\text{J/K}$. The change in entropy of the hot reservoir during the process $D$ is $-5.57\text{J/K}$.

Explanation of Solution

The number of moles of the diatomic gas is $1000\text{mol}$.

Write the formula for the net entropy change of the universe during one cycle.

$S=S_A+S_B+S_C+S_D$ (VI)

Here, $S$ is the net heat change in the cycle, $S_A$ is the entropy change of the universe in process $A$, $S_B$ is the entropy change of the universe during the process $B$, $S_C$ is the entropy change of the universe in process $C$, $S_D$ is the entropy change of the universe during process $D$.

Conclusion:

Substitute $-1.86\text{J/K}$ for $S_A$, $7.61\text{J/K}$ for $S_B$, $1.86\text{J/K}$ for $S_C$, $-5.57\text{J/K}$ for $S_D$.

$\begin{array}{c}S=-1.86\text{J/K}+7.61\text{J/K}+1.86\text{J/K}-5.57\text{J/K}\\ \text{=2}\text{.04}\text{J/K}\end{array}$

The net entropy change of the universe is $\text{2}\text{.04}\text{J/K}$.