91Ó°ÊÓ

a.

The ideal gas law,

$\mathrm{pV}=\mathrm{nRT}⇒\mathrm{T}=\frac{\mathrm{pV}}{\mathrm{nR}}$

The temperature at the second state is,

${\mathrm{T}}_2=\frac{4×{10}^5\mathrm{atm}×0.001{\mathrm{m}}^3}{0.02×8.314\mathrm{Î\copyright }}$

$=2400\mathrm{K}$

${\mathrm{V}}_1={\mathrm{V}}_3=:\mathrm{V}$

${\mathrm{p}}_2{\mathrm{V}}_2^{\mathrm{γ}}={\mathrm{p}}_1{\mathrm{V}}^{\mathrm{γ}}$

$⇒\mathrm{V}={\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{1/\mathrm{γ}}{\mathrm{V}}_2$

So,

$\mathrm{V}=4^{1/1.67}×0.001{\mathrm{m}}^3$

$=0.0023{\mathrm{m}}^3$

Because the pressure remains constant, the increase in temperature at state third will be equal to the increase in volume, as shown in the first expression. That is to say,

${\mathrm{T}}_3=2.3{\mathrm{T}}_2$

$=5520\mathrm{K}$

${\mathrm{T}}_1=0.25{\mathrm{T}}_3=1380\mathrm{K}$

b.

In process $1→2$, heat exchange is nil, whereas work done on the gas is equal to the change in internal energy, which is $\mathrm{W}=∆{\mathrm{E}}_{\mathrm{th}}$

$=0.02×20.8\mathrm{kJ}×1020\mathrm{K}=424\mathrm{J}$.

The work done on the gas in process $2→1$will be$0.02×29.1\mathrm{kJ}×3120\mathrm{K}=1820\mathrm{J}$.

The work done by the gas in process $3→1$will be zero, but the heat released will be equal to the drop in internal energy, which will be $-1720\mathrm{J}$.

Table is,

c.

The total net work done per cycle is $96\mathrm{J}$

The heat is $1820\mathrm{J}$.

The thermal efficiency is,

$\mathrm{η}=\frac{96\mathrm{J}}{1820\mathrm{J}}$

$=5.3\%$