Question

A heat engine using a monatomic gas follows the cycle shown in FIGURE P$21.58$.

a. Find ${\mathrm{W}}_{\mathrm{s}},\mathrm{Q}$, and ${\mathrm{\Delta E}}_{\text{th}}$for each process in the cycle. Display your results in a table.

b. What is the thermal efficiency of this heat engine?

Short answer

a. Table for process of cycle is .

b. The thermal efficiency is$15\%$.

Step-by-step solution

Calculation for work done in first process (part a)

(a) .

The third process is adiabatic,

So the maximal pressure is,

${\mathrm{p}}_1{\mathrm{V}}_1^{\mathrm{\gamma }}={\mathrm{p}}_3{\mathrm{V}}_3^{\mathrm{\gamma }}$

$\Rightarrow {\mathrm{p}}_1={\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_1}\right)}^{\mathrm{\gamma }}{\mathrm{p}}_3$

So,

${\mathrm{p}}_1=6^{1.67}\times 100\mathrm{atm}$

$=1993\mathrm{kPa}$

The first process is isobaric,

so,

${\mathrm{T}}_1=\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}{\mathrm{T}}_2$

$=100\mathrm{K}$

The heat is,

$\mathrm{Q}=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{RT}}_2}{\mathrm{C}}_{\mathrm{p}}(600-100){\mathrm{m}}^3$

$\approx 8300\mathrm{J}$

The work done is,

role="math" localid="1650269695241" $\mathrm{W}=1993\times {10}^3\mathrm{J}\times 500\times {10}^{-6}$

$\approx 1000\mathrm{J}$

Calculation for work in second process (part a) solution

a.

For second process the work done is null,

$\mathrm{Q}=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{RT}}_2}{\mathrm{C}}_{\mathrm{p}}\mathrm{\Delta T}$

$=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{RT}}_2}{\mathrm{C}}_{\mathrm{p}}\frac{{\mathrm{p}}_2-{\mathrm{p}}_3}{{\mathrm{p}}_2}{\mathrm{T}}_2$

$=\frac{{\mathrm{C}}_{\mathrm{p}}{\mathrm{V}}_2\left({\mathrm{p}}_2-{\mathrm{p}}_3\right)}{\mathrm{R}}$

So,

$\mathrm{Q}=\frac{12.5\times 6\times {10}^{-4}{\mathrm{m}}^3\times 1893\times {10}^3\mathrm{J}}{8.314\mathrm{\Omega }}$

$\approx 2850\mathrm{J}$

The third point is,

${\mathrm{T}}_3=\frac{{\mathrm{p}}_3}{{\mathrm{p}}_2}{\mathrm{T}}_2$

role="math" localid="1650270106364" $=\frac{100\mathrm{atm}}{1993\mathrm{kPa}}\times 600{\mathrm{m}}^3$

$=30.1\mathrm{K}$

The heat is,

$\mathrm{Q}=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{RT}}_2}{\mathrm{C}}_{\mathrm{v}}\left({\mathrm{T}}_3-{\mathrm{T}}_2\right)$

$=-210\mathrm{J}$

Result for process in cycle part(a) solution

a.

Calculation for thermal efficiency (part b)

b.

From the table,

The total heat input is $5240\mathrm{J}$,

The work done is $790\mathrm{J}$.

The efficiency is,

$\mathrm{\eta }=\frac{790\mathrm{J}}{5240\mathrm{J}}$

$=15\%$