Chapter 5.5, Problem 177RP

A tank with an internal volume of 1 m3 contains air at 800 kPa and 25°C. A valve on the tank is opened, allowing air to escape, and the pressure inside quickly drops to 150 kPa, at which point the valve is closed.  Assume there is negligible heat transfer from the tank to the air left in the tank.

1. (a)   Using the approximation he ≈ constant = he,avg = 0.5 (h1 + h2), calculate the mass withdrawn during the process.
2. (b)   Consider the same process but broken into two parts. That is, consider an intermediate state at P2 = 400 kPa, calculate the mass removed during the process from P1 = 800 kPa to P2 and then the mass removed during the process from P2 to P3 = 150 kPa, using the type of approximation used in part (a), and add the two to get the total mass removed.
3. (c)   Calculate the mass removed if the variation of he is accounted for.

FIGURE P5–185

To determine

The mass withdrawn during the process.

Answer to Problem 177RP

The mass withdrawn during the process is $6.618\text{kg}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (I)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the tank as the control volume. Initially the tank is filled with air and the valve is in closed position, further no other mass is allowed to enter the tank. Hence, the inlet mass is neglected i.e. $m_{\text{in}}=0$.

Rewrite the Equation (I) as follows.

$\begin{array}{l}0-m_{\text{e}}={\left(m_2-m_1\right)}_{\text{cv}}\\ -m_{\text{e}}=m_2-m_1\\ m_{\text{e}}=m_1-m_2\end{array}$ (II)

Write the formula for initial and final mass of air present in the tank.

$m_1=\frac{P_1\nu }{R{T}_1}$ (III)

$m_2=\frac{P_2\nu }{R{T}_2}$ (IV)

Here, the mass of air is $m$, the pressure is $P$, the volume is $\nu$, the gas constant of is $R$, the temperature is $T$, the subscript 1 indicates the initial state, and the subscript 2 indicates the final state.

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (V)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

When the valve is opened and air starts escape from the tank. Neglect the heat transfer and work done i.e. $W_{\text{in}}=W_e=0$ and $Q_{\text{in}}=Q_e=0$. Also neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$.

The Equation (V) reduced as follows.

$\begin{array}{l}0-m_{\text{e}}{h}_{\text{e}}=m_2{u}_2-m_1{u}_1\\ m_2{u}_2-m_1{u}_1+m_{\text{e}}{h}_{\text{e}}=0\end{array}$ (VI)

The enthalpy and internal energy in terms of temperature and specific heats are expressed as follows.

$\begin{array}{c}h=c_{p}T\\ u=c_{v}T\end{array}$

Rewrite the Equation (VI) as follows.

$m_2{c}_v{T}_2-m_1{c}_v{T}_1+\left(m_2-m_1\right)c_p{T}_e=0$ (VII)

The temperature of the air while exiting the tank is considered as the average temperature of initial and final temperatures.

$\begin{array}{c}{T}_e=\frac{T_1+T_2}{2}\\ =\frac{\left(25+273\right)\text{K}+T_2}{2}\\ =\frac{298\text{K}+T_2}{2}\\ =\frac{298+T_2}{2}\end{array}$

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of air is $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$.

Refer Table A-2b, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ of air corresponding to the ambient temperature is $1.005\text{kJ/kg}\cdot \text{K}$ and the specific heat at constant volume $\left(c_v\right)$ is $0.718\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $800\text{kPa}$ for $P_1$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $25\text{°C}$ for $T_1$ in Equation (III).

$\begin{array}{c}{m}_{\text{1}}=\frac{800\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(25°\text{C}\right)}\\ =\frac{800\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(25+273\right)}\\ =\frac{800\text{kPa}{\text{.m}}^3}{85.526\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}}\\ =9.3539\text{kg}\end{array}$

$\simeq 9.354\text{kg}$

Substitute $150\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, and $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{150\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(T_2\right)}\\ =\frac{522.6481}{T_2}\end{array}$

Substitute $\frac{522.6481}{T_2}$ for $m_2$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_v$, $9.354\text{kg}$ for $m_1$, $298\text{K}$ for $T_1$, $1.005\text{kJ/kg}\cdot \text{K}$ for $c_p$, and $\frac{298+T_2}{2}$ for $T_{\text{e}}$ in Equation (VII).

$\begin{array}{l}\left[\begin{array}{l}\left(\frac{522.6481}{T_2}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)T_2\\ -\left(9.354\text{kg}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)298\text{K}\\ +\left(9.354\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}375.2613\text{kJ}-2001.4192\text{kJ}\\ +\left(9.354\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}-1626.1579\text{kJ}\\ +\left(9.354\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left(9.354-\frac{522.6481}{T_2}\right)\left(1.005\right)\left(\frac{298+T_2}{2}\right)=1626.1579\end{array}$ (VIII)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (VIII) and obtain the value of $T_2$ and consider the positive root alone.

$\begin{array}{c}{T}_2=191.00864\text{K}\\ \simeq 191\text{K}\end{array}$

Substitute $150\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $191\text{K}$ for $T_2$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{150\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(191\text{K}\right)}\\ =\frac{150\text{kPa}\cdot {\text{m}}^3}{54.817\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =2.7364\text{kg}\end{array}$

Substitute $9.354\text{kg}$ for $m_1$ and $2.7364\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_{\text{e}}=9.354\text{kg}-2.7364\text{kg}\\ =6.6176\text{kg}\\ \simeq 6.618\text{kg}\end{array}$

Thus, the mass withdrawn during the process is $6.618\text{kg}$.

To determine

The mass withdrawn during the pressure reduced from $800\text{kPa}$ to $400\text{kPa}$ and $400\text{kPa}$ to $150\text{kPa}$.

Answer to Problem 177RP

The total mass withdrawn during the process 1-3 is $\text{6}\text{.551}\text{kg}$.

Explanation of Solution

Consider Process 1-2:

The pressure drop from $800\text{kPa}$ to $400\text{kPa}$.

$P_1=800\text{kPa,}{P}_2=400\text{kPa}$

Substitute $800\text{kPa}$ for $P_1$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $25\text{°C}$ for $T_1$ in Equation (III).

$\begin{array}{c}{m}_{\text{1}}=\frac{800\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(25°\text{C}\right)}\\ =\frac{800\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(25+273\right)}\\ =\frac{800\text{kPa}{\text{.m}}^3}{85.526\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}}\\ =9.3539\text{kg}\end{array}$

$\simeq 9.354\text{kg}$

Substitute $400\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, and $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{400\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(T_2\right)}\\ =\frac{1393.7282}{T_2}\end{array}$

Substitute $\frac{1393.7282}{T_2}$ for $m_2$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_v$, $9.354\text{kg}$ for $m_1$, $298\text{K}$ for $T_1$, $1.005\text{kJ/kg}\cdot \text{K}$ for $c_p$, and $\frac{298+T_2}{2}$ for $T_{\text{e}}$ in Equation (VII).

$\begin{array}{l}\left[\begin{array}{l}\left(\frac{1393.7282}{T_2}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)T_2\\ -\left(9.354\text{kg}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)298\text{K}\\ +\left(9.354\text{kg}-\frac{1393.7282}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}1000.6969\text{kJ}-2001.4192\text{kJ}\\ +\left(\frac{1393.7282}{T_2}-9.354\text{kg}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}-1000.7223\text{kJ}\\ +\left(\frac{1393.7282}{T_2}-9.354\text{kg}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{298+T_2}{2}\right)\end{array}\right]=0\\ \left(9.354-\frac{1393.7282}{T_2}\right)\left(1.005\right)\left(\frac{298+T_2}{2}\right)=1000.7223\end{array}$ (IX)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (IX) and obtain the value of $T_2$ and consider the positive root alone.

$\begin{array}{c}{T}_2=245.0751\text{K}\\ \simeq 245.1\text{K}\end{array}$

Substitute $400\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $245.1\text{K}$ for $T_2$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{400\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(245.1\text{K}\right)}\\ =\frac{400\text{kPa}\cdot {\text{m}}^3}{70.3437\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =5.6864\text{kg}\end{array}$

Substitute $9.354\text{kg}$ for $m_1$ and $5.6864\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_{\text{e}\left(1-2\right)}=9.354\text{kg}-5.6864\text{kg}\\ =3.6672\text{kg}\\ \simeq 3.667\text{kg}\end{array}$

Thus, the mass withdrawn during the process 1-2 is $3.667\text{kg}$.

Consider Process 2-3:

The pressure drop from $400\text{kPa}$ to $150\text{kPa}$.

$P_2=400\text{kPa,}{P}_3=150\text{kPa}$

Here, $T_1\cong T_2=245.1\text{K}$ and $T_3\cong T_2$.

Substitute $400\text{kPa}$ for $P_1$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $245.1\text{K}$ for $T_1$ in Equation (III).

$\begin{array}{c}{m}_{\text{1}}=\frac{400\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(245.1\text{K}\right)}\\ =\frac{400\text{kPa}\left(1{\text{m}}^3\right)}{70.3437}\\ =5.6864\text{kg}\end{array}$

$\simeq 9.354\text{kg}$

Substitute $150\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, and $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{150\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(T_2\right)}\\ =\frac{522.6481}{T_2}\end{array}$

Substitute $\frac{522.6481}{T_2}$ for $m_2$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_v$, $5.6864\text{kg}$ for $m_1$, $245.1\text{K}$ for $T_1$, $1.005\text{kJ/kg}\cdot \text{K}$ for $c_p$, and $\frac{245.1\text{K}+T_2}{2}$ for $T_{\text{e}}$ in Equation (VII).

$\begin{array}{l}\left[\begin{array}{l}\left(\frac{522.6481}{T_2}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)T_2\\ -\left(5.6864\text{kg}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)245.1\text{K}\\ +\left(5.6864\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{245.1+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}375.2613\text{kJ}-1000.7029\text{kJ}\\ +\left(5.6864\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{245.1+T_2}{2}\right)\end{array}\right]=0\\ \left[\begin{array}{l}-625.4416\text{kJ}\\ +\left(5.6864\text{kg}-\frac{522.6481}{T_2}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(\frac{245.1+T_2}{2}\right)\end{array}\right]=0\\ \left(5.6864-\frac{522.6481}{T_2}\right)\left(1.005\right)\left(\frac{245.1+T_2}{2}\right)=625.4416\end{array}$ (X)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (X) and obtain the value of $T_2$ and consider the positive root alone.

$\begin{array}{c}{T}_2=186.49217\text{K}\\ \simeq 186.5\text{K}\end{array}$

Substitute $150\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $186.5\text{K}$ for $T_2$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{150\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(186.5\text{K}\right)}\\ =\frac{150\text{kPa}\cdot {\text{m}}^3}{53.5255\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =2.8024\text{kg}\end{array}$

Substitute $5.6864\text{kg}$ for $m_1$ and $2.8024\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_{\text{e}\left(2-3\right)}=5.6864\text{kg}-2.8024\text{kg}\\ =2.884\text{kg}\end{array}$

Thus, the mass withdrawn during the process 2-3 is $2.884\text{kg}$.

The total mass withdrawn during the process 1-3 is as follows.

$\begin{array}{c}{m}_{\text{e}}=m_{\text{e}\left(1-2\right)}+m_{\text{e}\left(2-3\right)}\\ =3.667\text{kg}+2.884\text{kg}\\ =\text{6}\text{.551}\text{kg}\end{array}$

Thus, the total mass withdrawn during the process 1-3 is $\text{6}\text{.551}\text{kg}$.

To determine

The mass withdrawn during the process if there is variation in $h_e$.

Answer to Problem 177RP

The mass withdrawn during the process is $6.524\text{kg}$.

Explanation of Solution

Write the general mass balance equation.

$\begin{array}{l}{\dot{m}}_{\text{in}}-{\dot{m}}_{\text{e}}=\frac{d}{dt}\left(m_{\text{system}}\right)\\ {\dot{m}}_{\text{in}}-{\dot{m}}_{\text{e}}=\frac{d{m}_{\text{system}}}{dt}\end{array}$ (XI)

Here, the inlet mass flow rate is ${\dot{m}}_{\text{in}}$, the exit mass flow rate is ${\dot{m}}_{\text{e}}$, and the change in mass of the system is $\frac{d{m}_{\text{system}}}{dt}$.

Refer Equation (XI).

Write the mass balance equation for the given system.

$\begin{array}{l}{\dot{m}}_{\text{in}}-{\dot{m}}_{\text{e}}=\frac{dm}{dt}\\ 0-{\dot{m}}_{\text{e}}=\frac{dm}{dt}\\ \frac{dm}{dt}=-{\dot{m}}_{\text{e}}\end{array}$ (XII)

Rewrite the Equation (XII) as follows.

$\begin{array}{l}\frac{d\left(\frac{P\nu }{RT}\right)}{dt}=-{\dot{m}}_{\text{e}}\\ \frac{\nu }{R}\frac{d\left(P/T\right)}{dt}=-{\dot{m}}_{\text{e}}\end{array}$

Write the general energy rate balance equation.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (XIII)

Here, the rate of total energy in is ${\dot{E}}_{\text{in}}$, the rate of total energy out is ${\dot{E}}_{\text{out}}$, and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Refer Equation (XIII).

Write the energy balance equation for the given system.

$\begin{array}{l}\frac{d\left(mu\right)}{dt}+h_{\text{e}}{\dot{m}}_{\text{e}}=0\\ \frac{d\left(mu\right)}{dt}=-h_{\text{e}}{\dot{m}}_{\text{e}}\end{array}$ (XIV)

Here, the mass is $m$, the internal energy is $u$, and the enthalpy is $h$

Substitute $\frac{dm}{dt}$ for $-{\dot{m}}_{\text{e}}$ Equation (XIV).

$\frac{d\left(mu\right)}{dt}=h_{\text{e}}\left(\frac{dm}{dt}\right)$ (XV)

The enthalpy and internal energy is expressed as follows.

$\begin{array}{l}h=c_{p}T\\ u=c_{v}T\end{array}$

Substitute $c_p{T}_e$ for $h_e$ and $c_{v}T$ for $u$ in Equation (XV).

$\begin{array}{l}\frac{d\left(m{c}_{v}T\right)}{dt}=\left(c_{p}T\right)\left(\frac{dm}{dt}\right)\\ c_v\frac{d\left(mT\right)}{dt}=c_{p}T\left(\frac{dm}{dt}\right)\end{array}$ (XVI)

The mass of air in terms ideal gas is expressed as follows.

$m=\frac{P\nu }{RT}$

Rewrite the Equation (XVI) as follows.

$\begin{array}{l}{c}_v\frac{d\left(\frac{P\nu }{RT}T\right)}{dt}=c_{p}T\left(\frac{d\frac{P\nu }{RT}}{dt}\right)\\ c_v\frac{\nu }{R}\frac{dP}{dt}=c_{p}T\frac{\nu }{R}\frac{d\left(P/T\right)}{dt}\\ c_v\frac{dP}{dt}=c_{p}T\frac{d}{dt}\left(P/T\right)\end{array}$ (XVII)

Using $u/v$ differentiation rule expand $\frac{d}{dt}\left(P/T\right)$.

$\begin{array}{c}\frac{d}{dt}\left(P/T\right)=\frac{T\left(dP/dt\right)-P\left(dT/dt\right)}{T^2}\\ =\frac{T}{T^2}\left(\frac{dP}{dt}\right)-\frac{P}{T^2}\left(\frac{dT}{dt}\right)\\ =\frac{1}{T}\left(\frac{dP}{dt}\right)-\frac{P}{T^2}\left(\frac{dT}{dt}\right)\end{array}$

Substitute $\frac{1}{T}\left(\frac{dP}{dt}\right)-\frac{P}{T^2}\left(\frac{dT}{dt}\right)$ for $\frac{d}{dt}\left(P/T\right)$ in Equation (XVII).

$\begin{array}{l}{c}_v\frac{dP}{dt}=c_{p}T\left[\frac{1}{T}\left(\frac{dP}{dt}\right)-\frac{P}{T^2}\left(\frac{dT}{dt}\right)\right]\\ c_v\frac{dP}{dt}=c_p\left[\frac{T}{T}\left(\frac{dP}{dt}\right)-\frac{TP}{T^2}\left(\frac{dT}{dt}\right)\right]\\ c_v\frac{dP}{dt}=c_p\left[\left(\frac{dP}{dt}\right)-\frac{P}{T}\left(\frac{dT}{dt}\right)\right]\\ c_v\frac{dP}{dt}=c_p\left(\frac{dP}{dt}\right)-c_p\frac{P}{T}\left(\frac{dT}{dt}\right)\end{array}$

$\begin{array}{l}{c}_p\left(\frac{dP}{dt}\right)-c_v\frac{dP}{dt}=c_p\frac{P}{T}\left(\frac{dT}{dt}\right)\\ \left(c_p-c_v\right)\frac{dP}{dt}=c_p\frac{P}{T}\left(\frac{dT}{dt}\right)\\ \left(c_p-c_v\right)\frac{dP}{P}=c_p\frac{dT}{T}\\ \left(\frac{c_p-c_v}{c_p}\right)\frac{dP}{P}=\frac{dT}{T}\end{array}$

$\left(1-\frac{c_v}{c_p}\right)\frac{dP}{P}=\frac{dT}{T}$

Here, $k=\frac{c_p}{c_v}$.

$\begin{array}{l}\left(1-\frac{1}{k}\right)\frac{dP}{P}=\frac{dT}{T}\\ \left(\frac{k-1}{k}\right)\frac{dP}{P}=\frac{dT}{T}\end{array}$ (XVIII)

Integrate the Equation (XVIII) at the initial-1 and final-2 states.

$\begin{array}{l}\left(\frac{k-1}{k}\right){\displaystyle \underset{P_1}{\overset{P_2}{\int }}\frac{dP}{P}}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{dT}{T}}\\ \left(\frac{k-1}{k}\right){\left[\ln P\right]}_{P_1}^{P_2}={\left[\ln T\right]}_{T_1}^{T_2}\\ \left(\frac{k-1}{k}\right)\left[\ln P_2-\ln P_1\right]=\left[\ln T_2-\ln T_1\right]\\ \left(\frac{k-1}{k}\right)\ln \frac{P_2}{P_1}=\ln \frac{T_2}{T_1}\end{array}$

$\begin{array}{l}\ln {\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}=\ln \frac{T_2}{T_1}\\ {\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}=\frac{T_2}{T_1}\\ \frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}\end{array}$ (XIX)

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The specific heat ratio $\left(k\right)$ of air is $1.4$.

Conclusion:

Substitute $298\text{K}$ for $T_1$, $150\text{}\text{kPa}$ for $P_2$, $800\text{kPa}$ for $P_1$, and $1.4$ for $k$

$\begin{array}{l}\frac{T_2}{298\text{K}}={\left(\frac{150\text{}\text{kPa}}{800\text{kPa}}\right)}^{\frac{1.4-1}{1.4}}\\ T_2=298\text{K}{\left(0.1875\right)}^{0.2857}\\ T_2=184.7151\text{K}\end{array}$

Substitute $150\text{kPa}$ for $P_2$, $1{\text{m}}^3$ for $\nu$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $184.7151\text{K}$ for $T_2$ in Equation (III).

$\begin{array}{c}{m}_{\text{2}}=\frac{150\text{kPa}\left(1{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(184.7151\text{K}\right)}\\ =\frac{150\text{kPa}\cdot {\text{m}}^3}{53.5255\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =2.8295\text{kg}\\ =2.83\text{kg}\end{array}$

Substitute $9.354\text{kg}$ for $m_1$ and $2.83\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_{\text{e}}=9.354\text{kg}-2.83\text{kg}\\ =6.524\text{kg}\end{array}$

Thus, the mass withdrawn during the process is $6.524\text{kg}$.