Question

A 2.0 moldiatomic gas initially at 300 Kundergoes this cycle: It is (1) heated at constant volume to 800 K, (2) then allowed to expand isothermally to its initial pressure, (3) then compressed at constant pressure to its initial state. Assuming the gas molecules neither rotate nor oscillate, find (a) the net energy transferred as heat to the gas, (b) the net work done by the gas, and (c) the efficiency of the cycle.

Short answer

1. The net energy transferred as heat to the gas is$25.5\times {10}^3\mathrm{J}$.
2. The net work done by the gas is $4.73\times {10}^3\mathrm{J}$.
3. The efficiency of the cycle is $0.185$or $18.5\%$.

Step-by-step solution

The given data

The number of moles is $\mathrm{n}=2.00\mathrm{mol}$.

The initial temperature,$\mathrm{T}=300\mathrm{K}$.

The final temperature,$\mathrm{T}\text{'}=800\mathrm{K}$.

Process 1, heated at constant volume.

In process 2, isothermal expansion to its initial pressure.

In process 3, compressed at constant pressure to its initial state.

The gas molecules neither rotate nor oscillate.

Understanding the concept of thermodynamic relations

By using the formulas for input heat, the net work done and Equation 20-11, for the efficiency, we can find the value of net energy transferred as heat to the gas, the net work done by the gas, and the efficiency of the cycle respectively.

Formulae:

The work done in an isothermal process,${\mathrm{W}}_{\mathrm{iso}}=\mathrm{nRT}\text{'}\mathrm{In}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$ (1)

The energy is transferred as heat in the process at either constant pressure or constant volume,$\mathrm{Q}=\mathrm{nC}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}}\right)$ (2)

The efficiency of the cycle, $\mathrm{\epsilon }=\frac{\left|\mathrm{W}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}$ (3)

a) Calculation of net energy transferred as heat to the gas

Process 1 and 2 both require the input heat, ${\mathrm{Q}}_{\mathrm{H}}$. If the rotational degree of freedom is not involved, then $\gamma =\frac{5}{3}$and the molar-specific heat at constant volume and constant pressure are

$$\begin{gathered} {\mathrm{Q}}_2={\mathrm{W}}_2 \\ {\mathrm{C}}_{\mathrm{V}}=\frac{3}{2}\mathrm{R}\mathrm{and} \\ {\mathrm{C}}_{\mathrm{P}}=\frac{5}{2}\mathrm{R}\mathrm{respectively} \end{gathered}$$

Also, the volume ratio in process 2 is$\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}=\frac{8}{3}$

Since the working substance is an ideal gas, process 2 being isothermal implies

Therefore, the input heat ${\mathrm{Q}}_{\mathrm{H}}$is given using equations (1) and (2) as follows: (process 1 is a constant volume process and process 2 is an isothermal process)

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{H}}={\mathrm{Q}}_1+{\mathrm{Q}}_2 \\ ={\mathrm{nC}}_{\mathrm{V}}\left(\mathrm{T}\text{'}-\mathrm{T}\right)+\mathrm{nRT}\text{'}\mathrm{In}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ =2.00\mathrm{mol}\times \frac{3}{2}\times 8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}\times \left(800\mathrm{K}-300\mathrm{K}\right)+2.00\mathrm{mol}\times 8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 800\mathrm{K}\times \mathrm{In}\left(\frac{8}{3}\right) \\ =25.5\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the net energy transferred as heat is$25.5\times {10}^3\mathrm{J}$

b) Calculation of net work done by the gas

Using equation (2), the energy transferred as heat by the gas at constant pressure in process 3 is given as:

$$\begin{gathered} {\mathrm{Q}}_3=2.00\mathrm{mol}\times \frac{5}{2}\times 8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}\times \left(300\mathrm{K}-800\mathrm{K}\right) \\ =-20.8\times {10}^3\mathrm{J} \end{gathered}$$

So the net work done by the gas can be given as the sum of all energy transferred as heat as the change in internal energy of ideal gas is zero, that is

$$\begin{gathered} \mathrm{W}=25.5\times {10}^3\mathrm{J}-20.8\times {10}^3\mathrm{J} \\ =4.73\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of net work done by the gas is$4.73\times {10}^3\mathrm{J}$

c) Calculation of the efficiency of the cycle

Using the value of net work done by the gas and net energy of the gas in equation (3) we can get the efficiency of the cycle as given:

$$\begin{gathered} \mathrm{\epsilon }=\frac{4.73\times {10}^3\mathrm{J}}{25.5\times {10}^3\mathrm{J}} \\ \mathrm{\epsilon }=0.185 \\ \mathrm{\epsilon }\%=18.5\% \end{gathered}$$

Hence, the value of the efficiency of the cycle is$0.185$ or $18.5\%$