An ideal gas ( 1.0 mol) is the working substance in an engine that operates on the cycle shown in Figure 20-30. Processes BC andDA are reversible and adiabatic. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the engine efficiency?

The number of moles is n = 1.0 mol

Figure 20-30 is the graph of pressure vs. volume.

We use the concept of Ideal gas. Using the equation for the adiabatic process, we can find the value of gamma, and using the value of gamma, we can determine the type of gas. Using the equation of efficiency, we can find efficiency.

Formulae:

The equation of an adiabatic process for volume constant,

$P_i{V}_i^y=P_f{V}_f^y$ (1)

The efficiency of a Carnot engine,

$\epsilon =\frac{{\mathrm{Q}}_{\mathrm{in}}-{\mathrm{Q}}_{\mathrm{out}}}{{\mathrm{Q}}_{\mathrm{in}}}$ (2)

The heat transferred by the body,

$\mathrm{Q}=\mathrm{nC}∆\mathrm{T}$ (3)

The ideal-gas equation,

$\mathrm{PV}=\mathrm{nRT}$ (4)

We consider path D to A, using equation (1) and substituting the values, we can get that

$$\begin{gathered} {\mathrm{p}}_{\mathrm{D}}{\mathrm{V}}_{\mathrm{D}}^{\mathrm{y}}={\mathrm{p}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}^{\mathrm{y}} \\ \frac{{\mathrm{p}}_0}{32}{\left(8{\mathrm{V}}_0\right)}^{\mathrm{y}}={\mathrm{p}}_0{\mathrm{v}}_0^{\mathrm{y}} \\ {8}^{\mathrm{y}}=32 \\ \mathrm{\gamma }=\frac{\mathrm{In}\left(32\right)}{\mathrm{In}\left(8\right)} \\ \mathrm{\gamma }=\frac{5}{3} \end{gathered}$$

From this value, we can say that it is a monatomic gas.

For path A to B heat is absorbed, for a monatomic gas, ${\mathrm{C}}_{\mathrm{P}}=\frac{5}{2}\mathrm{R}$, we can get the heat transferred by the body using equation (3) and the given values as follows:

role="math" localid="1661574176097" $$\begin{gathered} {\mathrm{Q}}_{\mathrm{H}}=\mathrm{n}\left(\frac{5}{2}\mathrm{R}\right){\mathrm{T}}_{\mathrm{A}}\left(\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}-1\right) \\ {\mathrm{Q}}_{\mathrm{H}}={\mathrm{nRT}}_{\mathrm{A}}\left(\frac{5}{2}\right)(2-1) \\ fromequation\left(4\right),forcons\tan tpressure \\ \frac{T_B}{T_A}=\frac{V_B}{V_A} \\ \frac{T_B}{T_A}=\frac{2V_B}{V_A} \\ {Q}_H=p_0{V}_0\left(\frac{5}{2}\right) \end{gathered}$$

Similarly, we can write heat for path C to D as $\left(Q_L\right)$for lower temperature using equation (3) as given:

$$\begin{gathered} Q_L=n\left(\frac{5}{2}R\right)T_D\left(1-\frac{T_L}{T_D}\right) \\ =nR{T}_D\left(\frac{5}{2}\right)\left(1-2\right)..................\left(5\right) \end{gathered}$$

For point D, we can write

$$\begin{gathered} {\mathrm{nRT}}_{\mathrm{D}}=\frac{{\mathrm{p}}_0}{32}8{\mathrm{V}}_0 \\ =\frac{1}{4}{\mathrm{p}}_0{\mathrm{V}}_0 \end{gathered}$$

Using the above value, the equation (1) is given as:

$Q_L=-\frac{p_0{V}_0}{4}\left(\frac{5}{2}\right)$

The efficiency of the engine using equation (ii) is given as:

$$\begin{gathered} \mathrm{\epsilon }=1\frac{{\displaystyle \frac{{\mathrm{p}}_0{\mathrm{V}}_0}{4}}\left({\displaystyle \frac{5}{2}}\right)}{{\mathrm{p}}_0{\mathrm{V}}_0\left({\displaystyle \frac{5}{2}}\right)} \\ =1-\frac{1}{4} \\ =\frac{3}{4} \\ =0.75 \\ =75\% \end{gathered}$$

Hence, the value of the engine efficiency is 75%