91影视

a.

The ideal gas law,

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

$鈬�\mathrm{T}=\frac{\mathrm{pV}}{\mathrm{nR}}$

The temperature at the first state is,

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

$=600\mathrm{K}$

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

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

b.

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

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

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

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,

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

$=5.9\%$