A heat engine using $120\mathrm{mg}$of helium as the working substance follows the cycle shown in FIGURE P$21.61$.

a. Determine the pressure, temperature, and volume of the gas at points $1$,$2$, and $3$.

b. What is the engine's thermal efficiency?

c. What is the maximum possible efficiency of a heat engine that operates between ${\mathrm{T}}_{\max }$and${\mathrm{T}}_{\min }$?

a.

The number of moles is,

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

localid="1650282732447" $=\frac{0.120g}{4g}=0.03$

The ideal gas law,

$\mathrm{pV}=\mathrm{nRT}$

$\Rightarrow \mathrm{T}=\frac{\mathrm{pV}}{\mathrm{nR}}$

So,

localid="1650282936189" ${\mathrm{T}}_1=\frac{1\times {10}^5\mathrm{W}\times 1\times {10}^{-3}{\mathrm{m}}^3}{0.03\times 8.314\mathrm{\Omega }}$

$=400\mathrm{K}$

Although we could do the same for state two's temperature —

we don't need to do so for state three because it has the same temperature as state two because they are in the same isotherm —

we can also simply say that because the pressure was increased five times, the temperature was also increased five times, resulting in a temperature of

${\mathrm{T}}_2={\mathrm{T}}_3=2000\mathrm{K}$

The pressures in each of the three states are easily read as one,five and one atm, respectively.

State one and two have volumes of $1000$and $5000$cubic centimetres, respectively.

Given that process $2\to 3$ is isothermic, the volume at stage three is

${\mathrm{p}}_2{\mathrm{V}}_2={\mathrm{p}}_3{\mathrm{V}}_3$

role="math" localid="1650283234246" $\Rightarrow {\mathrm{V}}_3=\frac{{\mathrm{p}}_2}{{\mathrm{p}}_3}{\mathrm{V}}_2=5{\mathrm{V}}_2$

$=5000{\mathrm{cm}}^3$

b.

The work done in the first process was zero.

so,

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}$

localid="1650283488065" $=0.03\times 12.5\times (2000-400)\mathrm{J}$

$=600\mathrm{J}$

In the second process work is,

$\mathrm{W}=0.03\times 8.314\times 2000\ln \left(\frac{5000}{1000}\right)\mathrm{J}$

$=803\mathrm{J}$

In the third process, a work is,

$\mathrm{W}=1\times {10}^5\times (1-5)\times {10}^{-3}\mathrm{J}$

$=-400\mathrm{J}$

b.

Total the net work done per cycle is $803-400=403\mathrm{J}$.

The heat intake is localid="1651214783065" $2000\mathrm{J}$.

The efficiency is,

localid="1651214827693" $\mathrm{\eta }=\frac{403\mathrm{J}}{2000\mathrm{J}}=20.15\%$

c.

The utmost efficiency that an engine operating between the temperatures stated above might achieve would be a Carnot engine, with an efficiency of

${\mathrm{\eta }}_{\mathrm{C}}=\frac{2000-400\mathrm{J}}{2000\mathrm{J}}$

$=80\%$