A heat engine using a diatomic gas follows the cycle shown in FIGURE P$21.56$. Its temperature at point $1$is $20°\mathrm{C}$.

a. Determine ${\mathrm{W}}_{\mathrm{s}},\mathrm{Q}$, and ${\mathrm{\Delta E}}_{\text{th}}$for each of the three processes in this cycle. Display your results in a table.

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

c. What is the power output of the engine if it runs at $500\mathrm{rpm}$?

a.

The area beneath the surface can be used to determine the work in the first procedure.

It is going to be

${\mathrm{W}}_{\mathrm{s}}=\frac{30\times {10}^{-6}{\mathrm{cm}}^3(1.5-0.5)\mathrm{atm}\times {10}^5}{2}$

$=1.5\mathrm{J}$

The pressure was tripled and the volume was quadrupled in this procedure. This translates to a twelve-fold increase in temperature.

As a result, the temperature differential is eleven times greater than before. The change in internal energy can be calculated as follows:

${\mathrm{\Delta E}}_{\mathrm{th}}={\mathrm{nC}}_{\mathrm{v}}\mathrm{\Delta T}$

$=\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{RT}}_1}{\mathrm{C}}_{\mathrm{v}}\times 11{\mathrm{T}}_1$

role="math" localid="1650265225764" $=\frac{5\times {10}^4\mathrm{W}\times 1\times {10}^{-5}{\mathrm{m}}^3\times 11\times 20.8\mathrm{C}}{8.314\mathrm{\Omega }}$

$=13.75\mathrm{J}$

So,

$\mathrm{Q}={\mathrm{\Delta E}}_{\mathrm{th}}+{\mathrm{W}}_{\mathrm{s}}$

role="math" localid="1650265242221" $=13.75\mathrm{J}+1.5\mathrm{J}$

$=15.25\mathrm{J}$

Isochoric cooling is the second procedure, in which the pressure lowers by a third.

As a result, the temperature decreases by the same factor, resulting in an eight-fold change in temperature.

This means that the heat applied to the gas will be more intense.

$\mathrm{Q}=\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{nRT}}_1}{\mathrm{C}}_{\mathrm{v}}\times \left(-8{\mathrm{T}}_1\right)$

role="math" localid="1650265302070" $=\frac{5\times {10}^4\mathrm{W}\times 1\times {10}^{-5}\mathrm{m}\times (-8)\times 20.8\mathrm{C}}{8.314\mathrm{\Omega }}$

$=-10\mathrm{J}$

The change in internal energy will be the same because the work in the isochoric process is zero.

a.

In the third process we go isobarically from a temperature of four times is,

$\mathrm{Q}=\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{nRT}}_1}{\mathrm{C}}_{\mathrm{p}}\times \left(-4{\mathrm{T}}_1\right)$

$=\frac{5\times {10}^4\mathrm{W}\times 1\times {10}^{-5}\mathrm{m}\times (-4)\times 29\mathrm{C}}{8.314\mathrm{\Omega }}$

$=-7\mathrm{J}$

On the other hand, the work done on the other side can be represented by the rectangle under the graph. We can locate it right away.

${\mathrm{W}}_{\mathrm{s}}=-0.5\times {10}^5\times 30\times {10}^{-6}\mathrm{J}$

$=-3\mathrm{J}$

The change in internal energy is ,

${\mathrm{\Delta E}}_{\mathrm{th}}=3-7=-1\mathrm{J}$

b.

From the table,

The total heat input is $3.25\mathrm{J}$.

The work done is$1.5\mathrm{J}$.

The efficiency is,

role="math" localid="1650266315662" $\mathrm{\eta }=\frac{1.5\mathrm{J}}{3.25\mathrm{J}}$

$=46\%$

c.

work is $1.5\mathrm{J}$done.

Engine performs $500$cycles per minute,.

Power is,

role="math" localid="1650266459838" $\mathrm{P}=1.5\mathrm{J}\times \frac{500\mathrm{cyles}/\min }{60}$

$=12.5\mathrm{W}$