The heat engine shown in FIGURE P$21.62$uses $2.0\mathrm{mol}$of a monatomic gas as the working substance.

a. Determine ${\mathrm{T}}_1,{\mathrm{T}}_2$, and ${\mathrm{T}}_3$.

b. Make a table that shows ${\mathrm{\Delta E}}_{\text{th}},{\mathrm{W}}_{\mathrm{s}}$, and $\mathrm{Q}$for each of the three processes.

c. What is the engine's thermal efficiency?

a.

The ideal gas law,

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

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

The temperature at the first state is,

${\mathrm{T}}_1=\frac{4\times {10}^5\mathrm{Pa}\times 0.025{\mathrm{m}}^3}{2\times 8.314\mathrm{\Omega }}$

$=600\mathrm{K}$

${\mathrm{T}}_2=3{\mathrm{T}}_1=1800\mathrm{K}$

${\mathrm{T}}_3=2{\mathrm{T}}_1=1200\mathrm{K}$

b.

We have a job for process $1\to 2$that can be calculated as the area of the trapezoid, giving us $12.5\mathrm{kJ}$. localid="1650286493696" $∆{\mathrm{E}}_{\mathrm{th}}=2\times 12.5\mathrm{kJ}\times 1200\mathrm{K}=30\mathrm{J}$was discovered to be the energy change. The heat provided was$42.5\mathrm{J}$according to the first law.

For process $2\to 3$the work is zero, while the heat released is equal to the decrease in internal energy, both being $2\times 12.5\mathrm{kJ}\times 600\mathrm{K}=15\mathrm{kJ}$.

For process $3\to 1$the work done by the gas is $-10\mathrm{kJ}$The change in internal energy will be $2\times 12.5\mathrm{kJ}\times 600\mathrm{K}=15\mathrm{kJ}$.

The heat released will be$5\mathrm{kJ}$ .

b.

c.

The total work done per cycle is $2.5\mathrm{kJ}$.

The heat is $42.5\mathrm{kJ}$.

The thermal efficiency is,

$\mathrm{\eta }=\frac{2.5\mathrm{kJ}}{42.5\mathrm{kJ}}$

$=5.9\%$