Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

Answer to Problem 86P

The work done by engine during the 4 steps are $\text{692}\text{J}$,$\text{0}\text{J}$,$-\text{506}\text{J}$ ,$\text{0}\text{J}$ and the net work done is $\text{186}\text{J}$.

Explanation of Solution

Write the expression for work done during process A.

$W_A=nRT\ln \left(\frac{V_f}{V_i}\right)$

Here, T is the temperature, $W_A$ is the work done, $V_i$ is the volume, $V_f$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$W_C=nRT\ln \left(\frac{V_f}{V_i}\right)$

Here, T is the temperature, $W_C$ is the work done, $V_i$ is the volume, $V_f$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T,$0.020{\text{m}}^{\text{3}}$ for $V_i$, $0.025{\text{m}}^{\text{3}}$ for $V_f$, 1.00 mol for n and $8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R to get $W_A$.

$\begin{array}{c}{W}_A=\left(1.00\text{mol}\right)\left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(373\text{K}\right)\ln \left(\frac{0.025{\text{m}}^{\text{3}}}{0.020{\text{m}}^{\text{3}}}\right)\\ =\text{692}\text{J}\end{array}$

For the process C,

Substitute 273 K for T, $0.025{\text{m}}^{\text{3}}$ for $V_i$, $0.020{\text{m}}^{\text{3}}$ for $V_f$, 1.00 mol for n and $8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R to get $W_C$.

$\begin{array}{c}{W}_C=\left(1.00\text{mol}\right)\left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(273\text{K}\right)\ln \left(\frac{0.020{\text{m}}^{\text{3}}}{0.025{\text{m}}^{\text{3}}}\right)\\ =-\text{506}\text{J}\end{array}$

The net work done is,

$\begin{array}{c}{W}_{net}=\left(692\text{J}\right)+\left(0\text{J}\right)+\left(-506\text{J}\right)+\left(0\text{J}\right)\\ =\text{186}\text{J}\end{array}$

Therefore, work done by engine during the 4 steps are $\text{692}\text{J}$,$\text{0}\text{J}$,$-\text{506}\text{J}$ ,$\text{0}\text{J}$ and the net work done is $\text{186}\text{J}$.

(b)

To determine

The efficiency of the engine.

Answer to Problem 86P

The efficiency of the engine is $0.0670$.

Explanation of Solution

Write the expression for efficiency of the engine.

$e=\frac{W_{net}}{Q_{in}}$

Here, e is the efficiency and $Q_{in}$ is input heat.

Write the expression for input heat.

$Q_{in}=\frac{5}{2}nR\Delta T+W_A$

Here, $\Delta U$ is the change in internal energy and $\Delta T$ is the change intemperature.

Therefore, the input heat is,

$\begin{array}{c}{Q}_{in}=\frac{5}{2}\left(1.00\text{mol}\right)\left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(373\text{K}-273\text{K}\right)+\left(692\text{J}\right)\\ =2792\text{J}\end{array}$

Conclusion:

Substitute $\text{186}\text{J}$ for $W_{net}$ and $2792\text{J}$ for $Q_{in}$ to get e.

$\begin{array}{c}e=\frac{\text{186}\text{J}}{2792\text{J}}\\ =0.0670\end{array}$

Therefore, efficiency of the engine is $0.0670$.

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$.

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$e_r=1-\frac{T_C}{T_H}$

Here, $T_C$ is the temperature of cold reservoir and $T_H$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $T_C$ and 373 K for $T_H$ to get $e_r$.

$\begin{array}{c}{e}_r=1-\frac{\text{273}\text{K}}{\text{373}\text{K}}\\ =0.268\end{array}$

The ratio is given by,

$\begin{array}{c}\frac{e_r}{e}=\frac{0.268}{0.0670}\\ =4.00\end{array}$

The ratio of efficiency of reversible engine and the given engine is $4.00$.