91Ó°ÊÓ

a.

From the graph ,

${\mathrm{V}}_1={\mathrm{V}}_2=1000{\mathrm{cm}}^3$

The pressure at state two process $2→3$is adiabatic:

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

$⇒{\mathrm{p}}_2={\mathrm{p}}_3{\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right)}^{\mathrm{γ}}$

So,

${\mathrm{p}}_2=4^{1.4}×100\mathrm{atm}$

The temperature of the ideal gas law is,

$\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

$⇒{\mathrm{T}}_2=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{p}}_1{\mathrm{V}}_1}{\mathrm{T}}_1$

Numerically,

localid="1650272315505" ${\mathrm{T}}_2=\frac{696{\mathrm{Wm}}^3}{400{\mathrm{Wm}}^3}×300\mathrm{K}$

$=522\mathrm{K}$

b.

The ideal gas law,

${\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{nRT}}_1$

$⇒\mathrm{n}=\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{RT}}_1}$

$=\frac{4×{10}^5\mathrm{W}×1×{10}^{-3}{\mathrm{m}}^3}{8.314\mathrm{Î\copyright }×300\mathrm{K}}$

$=\underset{¯}{0.16}$

Process $1→2$:

Because the process is isochoric, the work done will be zero.

The heat will be equal to the change in internal energy, which will be equal to

$\mathrm{Q}={\mathrm{Δ·¡}}_{\mathrm{th}}$

$={\mathrm{nC}}_{\mathrm{V}}\mathrm{Δ°Õ}$

$=0.16×20.8{\mathrm{m}}^3×(522-300)\mathrm{K}$

$=739\mathrm{J}$

Process $2→3$:

The heat exchanged will be zero because the process is adiabatic.

The amount of effort done on the gas will be equal to the energy change.

$\mathrm{W}=-{\mathrm{Δ·¡}}_{\mathrm{th}}$

$=-{\mathrm{nC}}_{\mathrm{v}}\mathrm{Δ°Õ}$

$=-0.16×20.8{\mathrm{m}}^3×(300-522)\mathrm{K}$

$=739\mathrm{J}$

Process $3→1$

The change in internal energy will be zero because the process is isothermic.

The heat exchanged will be equal to the work minus the heat exchanged, implying that

$\mathrm{Q}=-\mathrm{W}=\mathrm{nRTln}\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_3}\right)$

$=0.16×8.314{\mathrm{m}}^3×300\mathrm{K}×\ln (0.25)$

$=-553\mathrm{J}$

b.

Table is,

c.

The total work done by the engine is,

${\mathrm{W}}_{\mathrm{e}}=-(-739+553)\mathrm{J}$

$=186\mathrm{J}$

The efficiency is,

$\mathrm{η}=\frac{186\mathrm{J}}{739\mathrm{J}}$

$=25\%$