Chapter 3, Problem 3.21P

(a)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=\text{1}\text{bar}$, $V_1=\text{12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ by mechanical reversible isothermal compression process, calculate values of $Q$, $W$, $\Delta U$ and $\Delta H$ .

Concept Introduction:

For isothermalcompressionprocess,

$PV=\text{constant}$

And

$W=-nR{T}_1\ln (\frac{V_2}{V_1})$

$\Delta U=0$

$Q=-W=nR{T}_1\ln (\frac{V_2}{V_1})$

$\Delta H=n{C}_P(T_2-T_1)=0$

Answer to Problem 3.21P

The values of values of $Q$, $W$, $\Delta U$ and $\Delta H$ for a process are:

$\Delta U=0$

$\Delta H=\text{0}$

$Q=\text{2982}\text{kJ}$

$W=\text{-2982}\text{kJ}$

Explanation of Solution

Given information:

It is given that $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=P_A\text{=1}\text{bar}$, $V_1=V_A\text{=12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ and $\gamma =\frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$

For isothermal compression process,

$\begin{array}{l}W=-nR{T}_1\ln (\frac{V_2}{V_1})=-P_1{V}_1\ln (\frac{P_1}{P_2})\\ W=1\text{bar×12}{\text{m}}^{\text{3}}\text{×ln(}\frac{\text{1}}{\text{12}}\text{)}\\ W=\text{29}\text{.82bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}\text{=2982}\text{kJ}\end{array}$

$Q=-W=\text{-2982}\text{kJ}$

“-” sign shows that heat is releasing, and volume is decreased on applying of work done on it.

(b)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=\text{1}\text{bar}$, $V_1=\text{12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ by mechanical reversible adiabatic compression followed by cooling at constant pressure process, calculate values of $Q$, $W$, $\Delta U$ and $\Delta H$ .

Concept Introduction:

The total amount of $Q$, $W$, $\Delta U$ and $\Delta H$ would be calculated using individual value of both process adiabatic compression followed by cooling at constant pressure. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1B adiabatic compressionprocess,

$P{V}^{\gamma }=\text{constant}$

And

$W=-\frac{P_1{V}_1-P_B{V}_B}{\gamma -1}$

$\Delta U=n{C}_V(T_2-T_1)=P_B{V}_B-P_1{V}_1$

$Q=0$

$\Delta H=\Delta U+(P_B{V}_B-P_1{V}_1)$

For step B2 constant pressureprocess,

$\frac{V_B}{T_B}=\frac{V_2}{T_2}$

And

$W=P(V_2-V_B)$

$\Delta U=n{C}_V(T_2-T_B)=P_2{V}_2-P_B{V}_B$

$Q=n{C}_P(T_2-T_B)=P_2{V}_2-P_B{V}_B$

$\Delta H=\Delta U+(P_2{V}_2-P_B{V}_B)$

Answer to Problem 3.21P

The values of values of $Q$, $W$, $\Delta U$ and $\Delta H$ for a process are:

$\Delta U=0$

$\Delta H=\text{0}$

$W_{12}=5105\text{kJ}$

$W=\text{-5105}\text{kJ}$

Explanation of Solution

Given information:

It is given that $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ and $P_1\text{=1}\text{bar}$, $V_1\text{=12}{\text{m}}^{\text{3}}$ to $P_2=P_B\text{=12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$, and $\gamma =\frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$ .

Since initial and final temperature is same in the process because $\begin{array}{l}{P}_1{V}_1=P_2{V}_2\\ 1\times 12=12\times 1\end{array}$ this implies $T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

$T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

So, we only need to do calculations for $Q$ and $W$ .

For step 1B adiabatic compressionprocess,

$\begin{array}{l}{P}_1{V}_1{}^{\gamma }=P_B{V}_B{}^{\gamma }\\ \end{array}$

$V_B=V_1{\left(\frac{P_1}{P_B}\right)}^{\frac{1}{\gamma }}$

$\begin{array}{l}{V}_B=12{\left(\frac{1}{12}\right)}^{\frac{3}{5}}\\ V_B=\text{2}\text{.702}{\text{m}}^{\text{3}}\end{array}$

$\begin{array}{l}{W}_{1B}=-\frac{P_1{V}_1-P_B{V}_B}{\gamma -1}=-\frac{1\times 12-12\times 2.702}{\frac{5}{3}-1}\\ W_{1B}=30.636\text{bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}=\text{3063}\text{kJ}\end{array}$

$Q_{1B}=-3063$

For step B2 constant pressure process,

$\begin{array}{l}{W}_{B2}=-P(V_2-V_B)=-12\times (1-2.702)\\ W_{B2}=20.424\text{bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}=2\text{042}\text{kJ}\end{array}$

$Q_{B2}=-2\text{042}\text{kJ}$

For whole process,

$\begin{array}{l}{Q}_{12}=Q_{1B}+Q_{B2}\\ Q_{12}=-3063-\text{2042}\\ Q_{12}=-5105\text{kJ}\end{array}$

$\begin{array}{l}{W}_{12}=W_{1B}+W_{B2}\\ W_{12}=3063+\text{2042}\\ W_{12}=5105\text{kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=\text{1}\text{bar}$, $V_1=\text{12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ by mechanical reversible adiabatic compression followed by cooling at constant volume process, calculate values of $Q$, $W$, $\Delta U$ and $\Delta H$ .

Concept Introduction:

The total amount of $Q$, $W$, $\Delta U$ and $\Delta H$ would be calculated using individual value of both process adiabatic compression followed by cooling at constant volume. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1C adiabatic compressionprocess,

$P{V}^{\gamma }=\text{constant}$

And

$W=-\frac{P_1{V}_1-P_C{V}_C}{\gamma -1}$

$\Delta U=n{C}_V(T_2-T_1)=P_C{V}_C-P_1{V}_1$

$Q=0$

$\Delta H=\Delta U+(P_C{V}_C-P_1{V}_1)$

For step C2 constant volume process,

$W=0$

$\Delta U=n{C}_V(T_2-T_B)=P_2{V}_2-P_C{V}_C$

$Q=n{C}_V(T_2-T_C)=P_2{V}_2-P_C{V}_C$

$\Delta H=\Delta U+(P_2{V}_2-P_C{V}_C)$

Answer to Problem 3.21P

The values of values of $Q$, $W$, $\Delta U$ and $\Delta H$ for a process are:

$\Delta U=0$

$\Delta H=\text{0}$

$W_{12}=7635\text{kJ}$

$Q_{12}=\text{-7635}\text{kJ}$

Explanation of Solution

Given information:

It is given that $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ and $P_1\text{=1}\text{bar}$, $V_1\text{=12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=V_C\text{=1}{\text{m}}^{\text{3}}$, and $\gamma =\frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$ .

Since initial and final temperature is same in the process because $\begin{array}{l}{P}_1{V}_1=P_2{V}_2\\ 1\times 12=12\times 1\end{array}$ this implies $T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

$T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

So, we only need to do calculations for $Q$ and $W$ .

For step 1Cadiabatic compressionprocess,

$\begin{array}{l}{P}_1{V}_1{}^{\gamma }=P_C{V}_C{}^{\gamma }\\ \end{array}$

$P_C=P_1{\left(\frac{V_1}{V_C}\right)}^{\gamma }$

$\begin{array}{l}{P}_C=1\times {\left(\frac{12}{1}\right)}^{\frac{5}{3}}\\ P_C=\text{62}\text{.898}\text{bar}\end{array}$

$\begin{array}{l}W=-\frac{P_1{V}_1-P_C{V}_C}{\gamma -1}=-\frac{1\times 12-1\times 62.898}{\frac{5}{3}-1}\\ W_{1C}=76.347\text{bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}=\text{7635}\text{kJ}\end{array}$

$Q_{1C}=-W=-7\text{635}\text{kJ}$

For step C2 constant volume process,

$\begin{array}{l}{W}_{C2}=0\\ Q_{C2}=0\end{array}$

For whole process,

$\begin{array}{l}{Q}_{12}=Q_{1C}+Q_{C2}\\ Q_{12}=-7635\text{kJ}\\ \end{array}$

$\begin{array}{l}{W}_{12}=W_{1C}+W_{C2}\\ W_{12}=7635+\text{0}\\ W_{12}=7635\text{kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=\text{1}\text{bar}$, $V_1=\text{12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ by mechanical reversible heating at constant volume followed by cooling at constant pressure, calculate values of $Q$, $W$, $\Delta U$ and $\Delta H$ .

Concept Introduction:

The total amount of $Q$, $W$, $\Delta U$ and $\Delta H$ would be calculated using individual value of both process. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1Dheating at constant volumeprocess,

$W=0$

$Q=P_D{V}_D-P_1{V}_1$

For step D2 cooling at constant pressureprocess,

$W=P(V_2-V_D)$

Answer to Problem 3.21P

The values of values of $Q$, $W$, $\Delta U$ and $\Delta H$ for a process are:

$\Delta U=0$

$\Delta H=\text{0}$

$W_{12}=13200\text{kJ}$

$Q_{12}=\text{-13200}\text{kJ}$

Explanation of Solution

Given information:

It is given that $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ and $P_1\text{=1}\text{bar}$, $V_1\text{=}{V}_D\text{=12}{\text{m}}^{\text{3}}$ to $P_2=P_D\text{=12}\text{bar}$, $V_2\text{=1}{\text{m}}^{\text{3}}$, and $\gamma =\frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$ .

Since initial and final temperature is same in the process because $\begin{array}{l}{P}_1{V}_1=P_2{V}_2\\ 1\times 12=12\times 1\end{array}$ this implies $T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

$T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

So, we only need to do calculations for $Q$ and $W$ .

For step 1D heating at constant volumeprocess

$\begin{array}{l}{W}_{1D}=0\\ Q_{1D}=0\end{array}$

For step D2 cooling at constant pressure process,

$W=-P(V_2-V_D)$

$\begin{array}{l}{W}_{D2}=12(V_D-V_2)=12\times (12-1)\\ W_{D2}=132\text{bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}=\text{13200}\text{kJ}\end{array}$

$Q_{D2}=-W_{D2}=-\text{13200}\text{kJ}$

For whole process,

$\begin{array}{l}{Q}_{12}=Q_{1D}+Q_{D2}\\ Q_{12}=0-1320\text{0}\\ Q_{12}=-13200\text{kJ}\end{array}$

$\begin{array}{l}{W}_{12}=W_{1D}+W_{D2}\\ W_{12}=0+13200\\ W_{12}=13200\text{kJ}\end{array}$

(e)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ is changed from $P_1=\text{1}\text{bar}$, $V_1=\text{12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=\text{1}{\text{m}}^{\text{3}}$ by mechanical reversible cooling at constant pressure followed by heating at constant volume, calculate values of $Q$, $W$, $\Delta U$ and $\Delta H$ .

Concept Introduction:

The total amount of $Q$, $W$, $\Delta U$ and $\Delta H$ would be calculated using individual value of both process. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1Ecooling at constant pressureprocess, volume decreases, heat will be released and negative and work done is positive.

$W=P(V_E-V_1)$

For step E2 heating at constant volumeprocess, pressure increases

$W=0$

Answer to Problem 3.21P

The values of values of $Q$, $W$, $\Delta U$ and $\Delta H$ for a process are:

$\Delta U=0$

$\Delta H=\text{0}$

$W_{12}=1100\text{kJ}$

$Q_{12}=\text{-1100}\text{kJ}$

Explanation of Solution

Given information It is given that $C_P=\frac{5}{\text{2}}\text{R}$, $C_V=\frac{3}{\text{2}}\text{R}$ and $P_1\text{=}{P}_E=\text{1}\text{bar}$, $V_1\text{=12}{\text{m}}^{\text{3}}$ to $P_2=\text{12}\text{bar}$, $V_2=V_E\text{=1}{\text{m}}^{\text{3}}$, and $\gamma =\frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$ .

Since initial and final temperature is same in the process because $\begin{array}{l}{P}_1{V}_1=P_2{V}_2\\ 1\times 12=12\times 1\end{array}$ this implies $T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

$T_1=T_2$

So $\Delta H_{12}=\text{0}$ and $\Delta U_{12}=0$ and $Q=-W$

So, we only need to do calculations for $Q$ and $W$ .

For step 1E cooling at constant pressureprocess,

$W=P(V_E-V_1)$

$\begin{array}{l}{W}_{1E}=-P(V_E-V_1)=-1\times (1-12)\\ W_{1E}=11\text{bar}{\text{.m}}^{\text{3}}\text{×}\frac{\text{101}\text{.325}\text{kPa}}{\text{1}\text{.01325}\text{bar}}=\text{1100}\text{kJ}\end{array}$

$Q_{1E}=-W=-\text{1100}\text{kJ}$

For step E2 heating at constant volume process,

$P_1\text{=}{P}_E=\text{1}\text{bar}$ and $V_2=V_E\text{=1}{\text{m}}^{\text{3}}$

$\begin{array}{l}{W}_{E2}=0\\ Q_{E2}=0\end{array}$

heating at constant volume causes increase in pressure inside vessel

For whole process,

$\begin{array}{l}{Q}_{12}=Q_{1E}+Q_{E2}\\ Q_{12}=-110\text{0}-\text{0}\\ Q_{12}=-1100\text{kJ}\end{array}$

$\begin{array}{l}{W}_{12}=W_{1E}+W_{E2}\\ W_{12}=1100+\text{0}\\ W_{12}=1100\text{kJ}\end{array}$