A heat engine using 1.0 mol of a monatomic gas follows the cycle shown in FIGURE P21.55. 3750 J of heat energy is transferred to the gas during process 1 -> 2.
a. Determine Ws, Q, and ΔE_th for each of the four processes in this cycle. Display your results in a table.
b. What is the thermal efficiency of this heat engine?

The number of moles the monatomic gas = 1 mol,
Pressure at point 1 and at point 4 = 300 kPA,
Heat transferred during process 1 to 2 = 3750 J
Initial temperature = 300 K.

From the graph we can see that the The heat engine follows a closed cycle with

process 1 → 2 and process 3 → 4 being isochoric and

process 2 → 3 and process 4 → 1 being isobaric

For isochoric process 1 → 2

W_s-12=0

Now calculate heat.

Q₁₂ = W_s-12 + ΔE_th-12

Substitute Ws =0 and find Q₁₂

Substitute Q₁₂ = 3750 J

ΔE_th-12 = Q₁₂ = 3750 J

Now find the temp at point 2.

$$\begin{gathered} Q_{1\to 2}=3750\mathrm{J}=\Delta E_{\text{th}}=n{C}_{\mathrm{V}}\Delta T \\ \Rightarrow \Delta T=\frac{3750\mathrm{J}}{n{C}_{\mathrm{V}}}=\frac{3750\mathrm{J}}{(1.0\mathrm{mol})\left(\frac{3}{2}R\right)}=\frac{3750\mathrm{J}}{(1.0\mathrm{mol})\left(\frac{3}{2}\right)(8.31\mathrm{J}/\mathrm{molK})}=301\mathrm{K} \\ \Rightarrow T_2-T_1=300.8\mathrm{K}\Rightarrow T_2=300.8\mathrm{K}+300\mathrm{K}=601\mathrm{K} \end{gathered}$$

Find the volume at point 2

$V_2=V_1=\frac{nR{T}_1}{p_1}=\frac{(1.0\mathrm{mol})(8.31\mathrm{J}/\mathrm{molK})(300\mathrm{K})}{3.0\times {10}^5\mathrm{Pa}}=8.31\times {10}^{-3}{\mathrm{m}}^3$

Find the pressure P₂ from the isochoric condition as below:
$\frac{p_2}{T_2}=\frac{p_1}{T_1}\Rightarrow p_2=\frac{T_2}{T_1}{p}_1=\left(\frac{601\mathrm{K}}{300\mathrm{K}}\right)\left(3.00\times {10}^5\mathrm{Pa}\right)=6.01\times {10}^5\mathrm{Pa}$

With the above values of P₂, V₂ and T₂, we can now obtain P₃, V₃ and T₃. as below

$$\begin{gathered} V_3=2V_2=1.662\times {10}^{-2}{\mathrm{m}}^3{p}_3=p_2=6.01\times {10}^5\mathrm{Pa} \\ \frac{T_3}{V_3}=\frac{T_2}{V_2}\Rightarrow T_3=\frac{V_3}{V_2}{T}_2=2T_2=1202\mathrm{K} \end{gathered}$$

For the isobaric process 2 → 3

$$\begin{gathered} Q_{2\to 3}=n{C}_{\mathrm{p}}\Delta T=(1.0\mathrm{mol})\left(\frac{5}{2}R\right)\left(T_3-T_2\right)=(1.0\mathrm{mol})\left(\frac{5}{2}\right)(8.31\mathrm{J}/\mathrm{molK})(601\mathrm{K})=12,480\mathrm{J} \\ {W}_{\mathrm{S}2\to 3}=p_3\left(V_3-V_2\right)=\left(6.01\times {10}^5\mathrm{Pa}\right)\left(8.31\times {10}^{-3}{\mathrm{m}}^3\right)=4990\mathrm{J} \\ \Delta E_{\text{th}2\to 3}=Q_{2\to 3}-W_{\mathrm{S}2\to 3}=12,480\mathrm{J}-4990\mathrm{J}=7490\mathrm{J} \end{gathered}$$

We are now able to obtain p₄, V₄ and T₄. We have

$$\begin{gathered} V_4=V_3=1.662\times {10}^{-2}{\mathrm{m}}^3{p}_4=p_1=3.00\times {10}^5\mathrm{Pa} \\ \frac{T_4}{p_4}=\frac{T_3}{p_3}\Rightarrow T_4=\frac{p_4}{p_3}{T}_3=\left(\frac{3.00\times {10}^5\mathrm{Pa}}{6.01\times {10}^5\mathrm{Pa}}\right)(1202\mathrm{K})=600\mathrm{K} \end{gathered}$$

For isochoric process 3 → 4

role="math" localid="1650384461160" $$\begin{gathered} Q_{3\to 4}=n{C}_{\mathrm{V}}\Delta T=(1.0\mathrm{mol})\left(\frac{3}{2}R\right)\left(T_4-T_3\right)=(1.0\mathrm{mol})\left(\frac{3}{2}\right)(8.31\mathrm{J}/\mathrm{molK})(-602)=-7500\mathrm{J} \\ {W}_{\mathrm{S}3\to 4}=0\mathrm{J}\Rightarrow \Delta E_{\text{th}3\to 4}=Q_{3\to 4}-W_{\mathrm{S}3\to 4}=-7500\mathrm{J} \end{gathered}$$

For isobaric process 4 → 1

$$\begin{gathered} Q_{4\to 1}=n{C}_{\mathrm{P}}\Delta T=(1.0\mathrm{mol})\frac{5}{2}(8.31\mathrm{J}/\mathrm{molK})(300\mathrm{K}-600\mathrm{K})=-6230\mathrm{J} \\ {W}_{\mathrm{S}4\to 1}=p_4\left(V_1-V_4\right)=\left(3.00\times {10}^5\mathrm{Pa}\right)\times \left(8.31\times {10}^{-3}{\mathrm{m}}^3-1.662\times {10}^{-2}{\mathrm{m}}^3\right)=-2490\mathrm{J} \\ \Delta E_{\text{th}4\to 1}=Q_{4\to 1}-W_{\mathrm{S}4\to 1}=-6230\mathrm{J}-(-2490\mathrm{J})=-3740\mathrm{J} \end{gathered}$$

Summarize in table

The number of moles the monatomic gas = 1 mol,
Pressure at point 1 and at point 4 = 300 kPA,
Heat transferred during process 1 to 2 = 3750 J
Initial temperature = 300 K.

And the diagram is given for the cycle.

The efficiency can be calculated as below

$\eta =\frac{W_{\text{out}}}{Q_{\mathrm{H}}}=\frac{W_{\text{out}}}{Q_{1\to 2}+Q_{2\to 3}}=\frac{2500\mathrm{J}}{3750\mathrm{J}+12,480\mathrm{J}}=0.154=15.4\%$