Chapter 15, Problem 100P

(a)

To determine

The number of moles of gas used in the engine.

Answer to Problem 100P

The number of moles of gas used in the engine is $13.0\text{mol}$.

Explanation of Solution

Find an expression for the number of moles.

$PV=nRT$ (I)

Here, $n$ is the number of moles, $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $R$ is the gas constant.

Rewrite the equation (I) for $n$.

$n=\frac{PV}{RT}$ (II)

Conclusion:

Substitute $1.1\text{atm}$ for $P$, $0.500{\text{m}}^3$ for $V$, $470.0\text{K}$ for $T$ and $8.314\text{J/mol}\cdot \text{K}$ for $R$ in equation (II) to find $n$.

$\begin{array}{c}n=\frac{\left(1.1\text{atm}\right)\left(0.500{\text{m}}^3\right)}{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(470.0\text{K}\right)}\\ =\frac{\left(\left(1.1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right)\left(0.500{\text{m}}^3\right)}{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(470.0\text{K}\right)}\\ =13.0\text{mol}\end{array}$

Thus, the number of moles of gas used in the engine is $13.0\text{mol}$.

(b)

To determine

The heat flow in the steps AB and CA.

Answer to Problem 100P

The heat flow in the step AB is $304\text{kJ}$ into the gas, and the heat flow in the step CA is $-380\text{kJ}$ out of the gas.

Explanation of Solution

Find an expression to find the heat flow in the step AB.

$\begin{array}{c}Q=n{C}_V\Delta T\\ =n{C}_V\left(T_f-T_i\right)\end{array}$ (III)

Here, $C_V$ is the specific heat at constant volume, $n$ is the number of moles, $\Delta T$ is the change in temperature, $T_f$ is the final temperature and $T_i$ is the initial temperature.

Substitute $\frac{3}{2}R$ for $C_V$, $\frac{P_{f}V}{nR}$ for $T_f$ and $\frac{P_{i}V}{nR}$ for $T_i$ in equation (II) to find $Q_{AB}$.

$\begin{array}{c}{Q}_{AB}=n\left(\frac{3}{2}R\right)\left(\frac{P_{f}V}{nR}-\frac{P_{i}V}{nR}\right)\\ =\frac{3V}{2}\left(P_f-P_i\right)\end{array}$ (IV)

Substitute $\frac{3}{2}R$ for $C_V$, $\frac{P{V}_f}{nR}$ for $T_f$ and $\frac{P{V}_i}{nR}$ for $T_i$ in equation (III) to find $\Delta U_{CA}$.

$\begin{array}{c}\Delta U_{CA}=n\left(\frac{3}{2}R\right)\left(\frac{P{V}_f}{nR}-\frac{P{V}_i}{nR}\right)\\ =\frac{3P}{2}\left(V_f-V_i\right)\end{array}$ (V)

Here, $V_f$ is the final volume and $V_i$ is the initial volume.

Find an expression to find the output work done.

$\begin{array}{c}{W}_{CA}=P\Delta V\\ =P\left(V_f-V_i\right)\end{array}$ (VI)

Here, $W_{CA}$ is the input work done.

Find an expression to find the heat flow in the step CA.

$Q_{CA}=\Delta U_{CA}-W_{CA}$ (VII)

Here, $Q_{CA}$ is the heat flow in the step CA.

Conclusion:

Substitute $0.500{\text{m}}^3$ for $V$, $5.00\text{atm}$ for $P_f$ and $1.00\text{atm}$ for $P_i$ in equation (IV) to find $Q_{AB}$

$\begin{array}{c}{Q}_{AB}=\frac{3\left(0.500{\text{m}}^3\right)}{2}\left(5.00\text{atm}-1.00\text{atm}\right)\\ =\frac{3\left(0.500{\text{m}}^3\right)}{2}\left(5.00\text{atm}-1.00\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\\ =\left(303,900\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =304\text{kJ}\end{array}$

Substitute $1.00\text{atm}$ for $P$, $0.500{\text{m}}^3$ for $V_f$ and $2.00{\text{m}}^3$ for $V_i$ in equation (V) to find $\Delta U_{CA}$

$\begin{array}{c}\Delta U_{CA}=\frac{3\left(1.00\text{atm}\right)}{2}\left(0.500{\text{m}}^3-2.00{\text{m}}^3\right)\\ =\frac{3\left(1.00\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)}{2}\left(0.500{\text{m}}^3-2.00{\text{m}}^3\right)\\ =\left(-227,925\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =-228\text{kJ}\end{array}$

Substitute $1.00\text{atm}$ for $P$, $0.500{\text{m}}^3$ for $V_f$ and $2.00{\text{m}}^3$ for $V_i$ in equation (VI) to find $\Delta U_{CA}$.

$\begin{array}{c}{W}_{CA}=\left(1.00\text{atm}\right)\left(0.500{\text{m}}^3-2.00{\text{m}}^3\right)\\ =\left(1.00\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(0.500{\text{m}}^3-2.00{\text{m}}^3\right)\\ =\left(-151,950\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =-152\text{kJ}\end{array}$ (VIII)

Substitute $-152\text{kJ}$ for $W_{CA}$ and $-228\text{kJ}$ for $\Delta U_{CA}$ to find $Q_{CA}$.

$\begin{array}{c}{Q}_{CA}=-228\text{kJ}-152\text{kJ}\\ \text{=}-\text{380}\text{kJ}\end{array}$

Thus, the heat flow in the step AB is $304\text{kJ}$ into the gas, and the heat flow in the step CA is $-380\text{kJ}$ out of the gas.

(c)

To determine

The work done in each step.

Answer to Problem 100P

The work done in the step AB is $0$, the work done in the step CA is $-152\text{kJ}$ and the work done in the step BC is $456\text{kJ}$.

Explanation of Solution

Find an expression to find the net work done on the system.

$W_{net}=-\frac{1}{2}\Delta P\Delta V$ (IX)

Here, $W_{net}$ is the net work done on the system.

Find an expression to find the work done in step BC.

$W_{BC}=W_{net}-\left(W_{AB}+W_{CA}\right)$ (X)

Here, $W_{BC}$ is the work done in step BC.

Conclusion:

The work done in the step AB is zero as no work is done in a isochoric process.

The work done in the step CA is $-152\text{kJ}$ as found in equation (VII).

Substitute $\left(5.00\text{atm}-1.00\text{atm}\right)$ for $\Delta P$ and $\left(2.00{\text{m}}^3-0.500{\text{m}}^3\right)$ for $\Delta V$ in equation (IX) to find $W_{net}$.

$\begin{array}{c}{W}_{net}=-\frac{1}{2}\left(5.00\text{atm}-1.00\text{atm}\right)\left(2.00{\text{m}}^3-0.500{\text{m}}^3\right)\\ =-\frac{1}{2}\left(5.00\text{atm}-1.00\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(2.00{\text{m}}^3-0.500{\text{m}}^3\right)\\ =-\left(303,900\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =-304\text{kJ}\end{array}$

Substitute $-304\text{kJ}$ for $W_{net}$, $0$ for $W_{AB}$ and $152\text{kJ}$ for $W_{CA}$ in equation (X) to find $W_{BC}$.

$\begin{array}{c}{W}_{BC}=-304\text{kJ}-\left(0+152\text{kJ}\right)\\ =-304\text{kJ}-152\text{kJ}\\ =-456\text{kJ}\end{array}$

Thus, the work done in the step AB is $0$, the work done in the step CA is $-152\text{kJ}$ and the work done in the step BC is $456\text{kJ}$.