Question

Suppose 1.0 molof a monatomic ideal gas initially at 10 Land 300 K is heated at constant volume to 600 K, allowed to expand isothermally to its initial pressure, and finally compressed at constant pressure to its original volume, pressure, and temperature. During the cycle, what are (a) the net energy entering the system (the gas) as heat and (b) the net work done by the gas? (c) What is the efficiency of the cycle?

Short answer

1. The net energyentersthe system as heat is$7.2\times {10}^3\mathrm{J}$.
2. The net work done by the gas is$9.63\mathrm{x}{10}^2\mathrm{J}$.
3. The efficiency of the cycle is 13%.

Step-by-step solution

The given data

The number of moles of an ideal gas, n = 1.0 mol

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

The final temperature,$T_H=600K$

The initial volume of a gas,$V_i=10L$

Understanding the concept of thermodynamic equations

The net heat energy entering the system can be calculated by rearranging the equation of the first law of thermodynamics and substituting the equations for internal energy, and work done by the gas in an isothermal process. We can calculate the efficiency of the cycle using the equation of efficiency.

Formulae:

The change in internal energy of gas using the first law of thermodynamics,

${\mathrm{dE}}_{\mathrm{int}}=\mathrm{Q}-\mathrm{W}$ (1)

The work done per cycle for the isothermal process,

${\mathrm{W}}_{\mathrm{iso}}=\mathrm{nRTln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)$ (2)

The efficiency of the cycle, $\epsilon =\frac{W}{Q_H}$ (3)

The ideal-gas equation, PV = nRT (4)

The energy is transferred as heat at a constant volume, $\mathrm{Q}={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}$ (5)

The work done by the gas at constant pressure, $\mathrm{W}=\mathrm{P}∆\mathrm{V}$ (6)

(a) Calculation of the net energy entering as heat

The internal energy gets added to the system in the constant volume process and thus, it is given by substituting the value of equation (5) in equation (1) as given:

$$\begin{gathered} {\mathrm{dE}}_{\mathrm{int}}={\mathrm{Q}}_{\mathrm{v}} \\ ={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T} \end{gathered}$$

Also, the work is done by the gas, when it undergoes the isothermal process, so it is given by equation (2).

Here, T = 600 K , as the isothermal process happens at 600 K

Substituting these equations in equation (1), the equation$Q_H$will become

${\mathrm{Q}}_{\mathrm{H}}={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}+\mathrm{nRTln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)..........................................\left(7\right)$

For the ideal monatomic gas,

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

And, using the gas law in ratio form, we can write

$$\begin{gathered} \frac{V_f}{V_i}=\frac{T_H}{T_L} \\ =\frac{600K}{300K} \\ =2 \end{gathered}$$

Substituting the above values and given data in equation (7), we can get the energy entering as heat as given:

$$\begin{gathered} Q_H=nR\left[\frac{3}{2}∆T+T_H\ln \left(2\right)\right] \\ =1.0\times 8.314J/mol·K\left[\frac{3}{2}\left(600K-300K\right)+600K\times \ln \left(2\right)\right] \\ =7.2\times {10}^{3}J \end{gathered}$$

Hence, the value of the net energy entering as heat is$7.2\times {10}^{3}J$

(b) Calculation of the net work done by the gas

The net work done by the gas is the sum of work done during the isothermal process and the work done during the constant pressure process. So, it is given using equation (2) and equation (6), and the given values as:

$$\begin{gathered} \mathrm{W}={\mathrm{nRT}}_{\mathrm{H}}\ln \left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)+{\mathrm{P}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}\left(1-\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ ={\mathrm{RT}}_{\mathrm{H}}\ln \left(2\right)+{\mathrm{nRT}}_{\mathrm{L}}\left(1-2\right) \\ =\mathrm{nR}\left({\mathrm{T}}_{\mathrm{H}}\ln \left(2\right)-{\mathrm{T}}_{\mathrm{L}}\right) \\ =1.0\times 8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\left[600\mathrm{K}\times \ln \left(2\right)-300\mathrm{K}\right] \\ =9.63\times {10}^2\hspace{0.17em}\mathrm{J} \end{gathered}$$

Hence, the net work done by the gas is$9.63\times {10}^2\hspace{0.17em}\mathrm{J}$

(c) Calculation of the efficiency of the cycle

So, using the values from part A and part B in equation (3), we get the efficiency of the cycle as given:

$\begin{array}{ll}\epsilon =\left(\frac{9.63\times {10}^{2}J}{7.2\times {10}^{2}J}\right)& 100\%\\ =\left(0.13\right)\times 100\%& \\ =13\%& \end{array}$

Hence, the value of the efficiency is 13%