91Ó°ÊÓ

The number of moles is$\mathrm{n}=3.4\mathrm{mol}$.

The gas is a diatomic gas.

The temperature of the gas increased from${\mathrm{T}}_{\mathrm{L}}=200\mathrm{K}$to${\mathrm{T}}_{\mathrm{H}}=500\mathrm{K}$at constant volume.

The gas is isothermally expanded to its original pressure

The gas is contracted at constant pressure back to its original volume.

The molecules rotate but do not oscillate.

By using the equation for the isothermal expansion given by Equation (19-14) and the constant pressure compression given by Equation (19-48), we can find the net work done ${\mathbf{W}}_{\mathbf{net}}$. By usingEquation (19-14), we can find the input heat, ${\mathbf{Q}}_{\mathbf{in}}$. Finally, by calculating the ratio,$\frac{{\mathbf{W}}_{\mathbf{net}}}{{\mathbf{Q}}_{\mathbf{in}}}$ , we can find theefficiency of the given cycle.

Formulae:

The isothermal expansion given by Equation (19-14),$\mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)$ (1)

The constant pressure compression given by Equation (19-48), $\mathrm{W}=\mathrm{p}∆\mathrm{V}=\mathrm{nR}∆\mathrm{T}$ (2)

The net work done of a cycle, role="math" localid="1661326162648" ${\mathrm{W}}_{\mathrm{net}}={\mathrm{nRT}}_{\mathrm{H}}\mathrm{In}\left(\frac{{\mathrm{V}}_{\max }}{{\mathrm{V}}_{\min }}\right)+\mathrm{nR}\left({\mathrm{T}}_{\mathrm{L}}-{\mathrm{T}}_{\mathrm{H}}\right)$ (3)

The energy transferred as heat during the process,

${\mathrm{Q}}_{\mathrm{in}}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}\right)+{\mathrm{nRT}}_{\mathrm{H}}\mathrm{In}\left(\frac{{\mathrm{V}}_{\max }}{{\mathrm{V}}_{\min }}\right)$ (4)

The efficiency of the cycle, $\mathrm{Î\mu }=\frac{{\mathrm{W}}_{\mathrm{net}}}{{\mathrm{Q}}_{\mathrm{in}}}$ (5)

The net work done is figured from the isothermal expansion given by Equation (19-14) and the constant pressure compression given by Equation (19-48).

The isothermal expansion is given by Equation (1) and the constant pressure compression is given by Equation (2). Thus, the net work done of the gas using equation (3) can be calculated as:(where,${\mathrm{T}}_{\mathrm{H}}$is high temperature,${\mathrm{T}}_{\mathrm{L}}$is low temperature, and

role="math" localid="1661326536724" $$\begin{gathered} \frac{{\mathrm{V}}_{\max }}{{\mathrm{V}}_{\min }}=\frac{{\mathrm{T}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{L}}} \\ =\frac{5}{2} \end{gathered}$$)

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}=3.4×8.31\mathrm{J}/\mathrm{kg}·\mathrm{K}×500\mathrm{K}×\mathrm{In}\left(\frac{5}{2}\right)+3.4×8.31\mathrm{J}/\mathrm{kg}·\mathrm{K}×\left(200\mathrm{K}-500\mathrm{K}\right) \\ =4468\mathrm{J} \end{gathered}$$

Now, we identify the input heat as that transformed in steps 1 and 2. Thus, the input heat role="math" localid="1661326764086" ${\mathrm{Q}}_{\mathrm{in}}$is given using equation (4) as: (where, $C_v=\frac{5}{2}\mathrm{R}$, molar specific heat at constant volume.)

$$\begin{gathered} Q_{in}=Q_1+Q_2 \\ =3.4×\frac{5}{2}×8.31\mathrm{J}/\mathrm{kg}·\mathrm{K}\left(200\mathrm{K}-500\mathrm{K}\right)+3.4×8.31\mathrm{J}/\mathrm{kg}·\mathrm{K}×500\mathrm{K}×\mathrm{In}\left(\frac{5}{2}\right) \\ =34135\mathrm{J} \end{gathered}$$

Now, the efficiency of the cycle using equation (5) is given by:

$$\begin{gathered} \mathrm{Î\mu }=\frac{4468\mathrm{J}}{34135\mathrm{J}} \\ \mathrm{Î\mu }=0.131 \\ \mathrm{Î\mu }\%=13.1\% \end{gathered}$$

Hence, the value of efficiency is 0.131 or 13.1% .