Chapter 15, Problem 100QAP

To determine

To calculate:

The work done, absorbed heat, change in internal energy, and entropy change of an ideal gas of an engine.

Answer to Problem 100QAP

Thus, the work done, absorbed heat, change in internal energy, and entropy change of an ideal gas of an engine have calculated.

Explanation of Solution

Given data:

n =1.23 mol

Specific heat ratio, $\gamma$ =1.41

Temperature, T₁ =300K

Temperature, T₂ =600K

Pressure, P₁ =15 atm

Pressure, P₂ =3 atm

Formula used:

Work done is calculated by,

  $W=$

  $nR\left(\Delta T\right)$

Change in internal energy is calculated by,

  $\Delta U=n{C}_v\Delta T$

Heat absorbed is calculated by,

  $Q=n{C}_p\Delta T$

Change in entropy is calculated by,

  $\Delta S=\frac{Q}{\Delta T}$

Calculation:

Work done for each part of the cycle is,

  $\begin{array}{l}{W}_a=P_1(V_2-V_1)=P_1{V}_2-P_1{V}_1\\ \text{we know, }Pv=nRT\\ W_a=nR{T}_2-nR{T}_1=nR\left(T_2-T\right)\\ W_a=\left(1.23\right)\left(8.314\right)\left(600-300\right)\\ W_a=3.06kJ\\ W_b=nR{T}_2\ln \left(\frac{P_2}{P_1}\right)=\left(1.23\right)\left(8.314\right)\left(600\right)\ln \left(\frac{3}{15}\right)\\ W_b=-9.87kJ\end{array}$

  $\begin{array}{l}{W}_c=\left(1.23\right)\left(8.314\right)\left(300-600\right)\\ W_a=-3.06kJ\\ W_d=nR{T}_1\ln \left(\frac{P_2}{P_1}\right)=\left(1.23\right)\left(8.314\right)\left(300\right)\ln \left(\frac{3}{15}\right)\\ W_d=4.93kJ\end{array}$

Work done for complete cycle is,

  $\begin{array}{l}{W}_{total}=3.06-9.87-3.06+4.93\\ W_{total}=-4.94kJ\end{array}$

Absorbed heat for each part of the cycle is,

  $\begin{array}{l}Q=n{C}_p\Delta T\\ C_p=\frac{\gamma R}{\gamma -1}=\frac{(1.41)(8.314)}{1.41-1}=28.59=28.6\\ Q_a=(1.23)(28.6)(600-300)\\ Q_a=10.55kJ\\ Q_b=U+W_b=0+W_b\text{}\text{}\text{}\text{}\text{(isothermal process)}\\ Q_b=-9.87kJ\\ Q_c=(1.23)(28.6)(600-300)\\ Q_c=10.55kJ\\ Q_d=U+W_d=0+W_d\text{}\text{}\text{}\text{}\text{(isothermal process)}\\ Q_d=4.93kJ\end{array}$

Absorbed heat for complete cycle is,

  $\begin{array}{l}{Q}_{total}=10.55-9.87+10.55+4.93\\ Q_{total}=16.16kJ\end{array}$

Change in internal energy for each part of the cycle is,

  $\begin{array}{l}\Delta U=n{C}_V\Delta T\\ C_V=\frac{C_p}{\gamma }=\frac{28.6}{1.41}=20.28\end{array}$

  $\begin{array}{l}\Delta U_a=(1.23)(20.28)(600-300)\\ \Delta U_a=7.48kJ\\ \Delta U_b=0\text{(constant temperature, }\Delta T=0\text{)}\\ \Delta U_c=(1.23)(20.28)(300-600)\\ \Delta U_c=-7.48kJ\\ \Delta U_d=0\text{(constant temperature, }\Delta T=0\text{)}\end{array}$

Change in internal energy for complete cycle is,

  $\begin{array}{l}\Delta U_{total}=7.48k+0-7.48k+0\\ \Delta U_{total}=0J\end{array}$

Change in entropy for each part of th ecycel si,

  $\begin{array}{l}\Delta S=\frac{Q_a}{\Delta T}\\ \Delta S_a=\frac{10.55\times {10}^3}{600-300}=35.16J{K}^{-1}\\ \Delta S_b=0\text{ (constant temperature, }\Delta T=0\text{)}\\ \Delta S_c=\frac{10.55\times {10}^3}{300-600}=-35.16J{K}^{-1}\\ \Delta S_d=0\text{ (constant temperature, }\Delta T=0\text{)}\end{array}$

Change in entropy of for complete cycle is,

  $\begin{array}{l}\Delta S_{total}=35.16+0-35.16+0\\ \Delta S_{total}=0J{K}^{-1}\end{array}$

Conclusion:

Work done for complete cycle is, $W_{total}=-4.94kJ$

Absorbed heat for complete cycle is, ${\text{Q}}_{\text{total}}\text{=16}\text{.16kJ}$

Change in internal energy for complete cycle is, ${\text{ΔU}}_{\text{total}}\text{=0J}$

Change in entropy of for complete cycle is, ${\text{ΔS}}_{\text{total}}{\text{=0JK}}^{\text{-1}}$