91Ó°ÊÓ

$$\begin{gathered} \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{gas}\mathrm{is}\mathrm{n}=1.00\mathrm{mol} \\ \mathrm{Volume}\mathrm{at}\mathrm{C},{\mathrm{V}}_{\mathrm{c}}=8.00{\mathrm{V}}_{\mathrm{b}} \\ \mathrm{Pressure}\mathrm{at}\mathrm{B},{\mathrm{P}}_{\mathrm{b}}=10.0\mathrm{atm} \\ \mathrm{Volume}\mathrm{at}\mathrm{B},{\mathrm{V}}_{\mathrm{b}}=1.00×{10}^{-3}{\mathrm{m}}^3 \end{gathered}$$

We can use the formula of heat related to the number of moles, specific heat, and temperature. Using the ideal gas equation, we can find pressure. Also, we can use the second law of thermodynamics to find work. Using work and heat, we can find the efficiency.

Formulae:

The ideal-gas equation,

$\mathrm{PV}=\mathrm{nRT}$ (1)

The equation of an adiabatic process for volume constant,

$$\begin{gathered} {\mathrm{PV}}^{\mathrm{y}}=\mathrm{C} \\ {\mathrm{P}}^{1-\mathrm{y}}{\mathrm{T}}^{\mathrm{y}}=\mathrm{C} \\ {\mathrm{TV}}^{\mathrm{y}-1}=\mathrm{C} \end{gathered}$$ (2)

The efficiency of a Carnot engine,

$$\begin{gathered} \mathrm{Î\mu }=\frac{{\mathrm{Q}}_{\mathrm{in}}-{\mathrm{Q}}_{\mathrm{out}}}{{\mathrm{Q}}_{\mathrm{in}}} \\ \mathrm{Î\mu }=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{in}}} \end{gathered}$$ (3)

The heat transferred by the body,

$\mathrm{Q}=\mathrm{mc}∆\mathrm{T}$ (4)

The internal energy using the first law of thermodynamics,

$\mathrm{dE}=\mathrm{dQ}-\mathrm{dW}$ (5)

For the path a to b, the volume is constant, and the specific heat for monatomic gas is

${\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\mathrm{R}$

From equation (1), we can get the relation as given:

$$\begin{gathered} \left({\mathrm{p}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}-{\mathrm{p}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{b}}\right)=\mathrm{nR}∆\mathrm{T} \\ \left({\mathrm{p}}_{\mathrm{b}}-{\mathrm{p}}_{\mathrm{a}}\right){\mathrm{V}}_{\mathrm{b}}=\mathrm{nR}∆\mathrm{T} \\ ∆\mathrm{T}=\frac{\left({\mathrm{p}}_{\mathrm{b}}-{\mathrm{p}}_{\mathrm{a}}\right){\mathrm{V}}_{\mathrm{b}}}{\mathrm{nR}} \end{gathered}$$

Substituting the above values and given values in equation (4), we get the energy added to the gas as heat as follows:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{in}}=\mathrm{n}\left(\frac{3}{2}\mathrm{R}\right)\frac{\left({\mathrm{p}}_{\mathrm{b}}-{\mathrm{p}}_{\mathrm{a}}\right){\mathrm{V}}_{\mathrm{b}}}{\mathrm{nR}} \\ =\frac{3}{2}\left({\mathrm{p}}_{\mathrm{b}}-{\mathrm{p}}_{\mathrm{a}}\right){\mathrm{V}}_{\mathrm{b}}.................................\left(6\right) \end{gathered}$$

${\mathrm{p}}_{\mathrm{a}}$is same as ${\mathrm{p}}_{\mathrm{c}}$, and b to c is an adiabatic path, so using equation (2), we can write the pressure values as:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{c}}={\mathrm{p}}_{\mathrm{a}} \\ ={\mathrm{p}}_{\mathrm{b}}{\left(\frac{{\mathrm{V}}_{\mathrm{b}}}{{\mathrm{V}}_{\mathrm{c}}}\right)}^{\mathrm{y}} \end{gathered}$$

For a monatomic gas $\mathrm{y}=\frac{5}{3}$, thus the pressure at A is given by substituting the given values as:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{a}}=1.013×{10}^6\mathrm{Pa}{\left(\frac{1.00×{10}^{-3}{\mathrm{m}}^3}{8\left(1.00×{10}^{-3}{\mathrm{m}}^3\right)}\right)}^{\frac{5}{3}} \\ =3.165625×{10}^4 \\ ≈3.166×{10}^4\mathrm{Pa} \end{gathered}$$

Plugging this value in equation (6), we get the heat added to the gas as given:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{in}}=\frac{3}{2}\left(1.013×{10}^6\mathrm{Pa}-3.166×{10}^4\mathrm{Pa}\right)1.00×{10}^{-3}{\mathrm{m}}^3 \\ =1.472×{10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the energy added to the gas is $1.472×{10}^3\mathrm{J}$

For path c to a pressure is constant, so,

${\mathrm{C}}_{\mathrm{p}}=\frac{5}{2}\mathrm{R}\mathrm{and}{\mathrm{p}}_{\mathrm{a}}={\mathrm{p}}_{\mathrm{c}}$and we get, the temperature difference by using equation (1) as follows:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{a}}\left({\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{c}}\right)=\mathrm{nR}∆\mathrm{T} \\ ∆\mathrm{T}=\frac{{\mathrm{p}}_{\mathrm{a}}\left({\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{c}}\right)}{\mathrm{nR}} \end{gathered}$$

Using this value in equation (4), we get the expression for the energy leaving the gas as heat as given:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{out}}=\frac{\mathrm{n}5}{2}\mathrm{R}\frac{{\mathrm{p}}_{\mathrm{a}}\left({\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{c}}\right)}{\mathrm{nR}} \\ =\frac{5}{2}{\mathrm{p}}_{\mathrm{a}}\left({\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{c}}\right) \end{gathered}$$

From the graph,${\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}$plugging the values in the above equation, we get

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{out}}=\frac{5}{2}\left(3.166×{10}^4\mathrm{Pa}\right)\left(1.00×{10}^{-3}{\mathrm{m}}^3-8\left(1.00×{10}^{-3}{\mathrm{m}}^3\right)\right) \\ =\frac{5}{2}\left(3.166×{10}^4\mathrm{Pa}\right)\left(-7.00\left(1.00×{10}^{-3}{\mathrm{m}}^3\right)\right) \\ =-5.54×{10}^2\mathrm{J} \\ \left|{\mathrm{Q}}_{\mathrm{out}}\right|=5.54×{10}^2\mathrm{J} \\ \mathrm{Hence},\mathrm{the}\mathrm{energy}\mathrm{leaving}\mathrm{the}\mathrm{gas}\mathrm{as}\mathrm{heat}\mathrm{is}5.54×{10}^2\mathrm{J} \end{gathered}$$

For a complete cycle, the change in internal energy of the system is zero, so we get the equation (5) as follows:

$$\begin{gathered} 0=\mathrm{dQ}-\mathrm{dW} \\ \mathrm{dW}=\mathrm{dQ} \\ \mathrm{dW}={\mathrm{Q}}_{\mathrm{in}}-\left|{\mathrm{Q}}_{\mathrm{out}}\right| \\ \mathrm{dW}=1.472×{10}^3\mathrm{J}-5.54×{10}^2\mathrm{J} \\ \mathrm{dW}=9.18×{10}^2\mathrm{J} \end{gathered}$$

Hence, the value of the net work done by the gas is$9.18×{10}^2\mathrm{J}$

Using equation (3) and the above-derived values, we can get the efficiency of the cycle

as given:

$$\begin{gathered} \mathrm{Î\mu }=\left(\frac{9.16×{10}^2\mathrm{J}}{1.467×{10}^3\mathrm{J}}\right)×100\% \\ =0.6244×100\% \\ =62.4\% \end{gathered}$$

Hence, the value of the efficiency of the gas is 62.4%