Question

A three-step cycle is undergone by 3.4mol of an ideal diatomic gas: (1) the temperature of the gas is increased from 200 K to 500 Kat constant volume; (2) the gas is then isothermally expanded to its original pressure; (3) the gas is then contracted at constant pressure back to its original volume. Throughout the cycle, the molecules rotate but do not oscillate. What is the efficiency of the cycle?

Short answer

The efficiency of the cycle is 0.131 or 13.1% .

Step-by-step solution

The given data

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.

Understanding the concept of thermodynamic relations

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, $\mathrm{\epsilon }=\frac{{\mathrm{W}}_{\mathrm{net}}}{{\mathrm{Q}}_{\mathrm{in}}}$ (5)

Calculation of the efficiency of the cycle

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

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

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.)

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

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

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

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