Chapter 13.3, Problem 93RP

A rigid tank contains a mixture of 4 kg of He and 8 kg of O₂ at 170 K and 7 MPa. Heat is now transferred to the tank, and the mixture temperature rises to 220 K. Treating the He as an ideal gas and the O₂ as a nonideal gas, determine (a) the final pressure of the mixture and (b) the heat transfer.

To determine

The final pressure of the mixture.

Answer to Problem 93RP

The final pressure of the mixture is $\text{9}\text{.97}\text{MPa}$.

Explanation of Solution

Write the expression to caculate the mole number of $\text{He}$ $\left(N_{\text{He}}\right)$.

$N_{\text{He}}=\frac{m_{\text{He}}}{M_{\text{He}}}$ (I)

Here, molar mass of $\text{He}$ is $M_{\text{He}}$ and mass of $\text{He}$ is $m_{\text{He}}$.

Write the expression to caculate the mole number of ${\text{O}}_2$ $\left(N_{{\text{O}}_2}\right)$.

$N_{{\text{O}}_2}=\frac{m_{{\text{O}}_2}}{M_{{\text{O}}_2}}$ (II)

Here, molar mass of ${\text{O}}_2$ is $M_{{\text{O}}_2}$ and mass of ${\text{O}}_2$ is $m_{{\text{O}}_2}$.

Write the expression to caculate the total number of moles $\left(N_m\right)$.

$N_m=N_{\text{He}}+N_{{\text{O}}_2}$ (III)

Write the expression to caculate the partial volume of helium $\left({\nu }_{\text{He}}\right)$.

${\nu }_{\text{He}}=\frac{N_{\text{He}}{R}_u{T}_1}{P_{m,1}}$ (IV)

Here, universal gas constant is $R_u$, temperature at state 1 is $T_1$, and pressure of the mixture is $P_m$.

Write the equation to calculate the reduced temperature and pressure at initial state.

$P_{R,1}=\frac{P_{m,1}}{P_{{\text{cr,O}}_2}}$ (V)

$T_{R,1}=\frac{T_1}{T_{{\text{cr,O}}_2}}$ (VI)

Write the expression to caculate the partial volume of helium $\left({\nu }_{{\text{O}}_2}\right)$.

${\nu }_{{\text{O}}_2}=\frac{Z{N}_{{\text{O}}_2}{R}_u{T}_1}{P_{m,1}}$ (VII)

Here, the compressibility factor is Z

Write the expression to calculate volume of the tank.

${\nu }_{\text{tank}}={\nu }_{\text{He}}+{\nu }_{{\text{O}}_2}$ (VIII)

Write the expression to calculate partial pressure at final stage.

$P_{\text{He,2}}=\frac{N_{\text{He}}{R}_u{T}_2}{{\nu }_{\text{tank}}}$ (IX)

Write the equation to calculate the reduced temperature of argon gas after mixing.

$T_{R_2,{\text{O}}_2}=\frac{T_2}{T_{{\text{cr,O}}_2}}$ (X)

Here, temperature at final stage is $T_2$ and critical temperature of oxygen is $T_{\text{cr,Ar}}$.

Write the equation to calculate the reduced volume of argon gas.

${\nu }_{R,{\text{O}}_2}=\frac{V_m/N_{{\text{O}}_2}}{R_u{T}_{\text{cr},{\text{O}}_2}/P_{\text{cr},{\text{O}}_2}}$ (XI)

Here, the mole number of oxygen gas $N_{{\text{O}}_2}$.

Write the equation to calculate the pressure of argon and nitrogen gases.

$P_{{\text{O}}_2}={\left(P_R{P}_{\text{cr}}\right)}_{{\text{O}}_2}$ (XII)

Write the expression to calculate the total final pressure.

$P_{m,2}=P_{\text{He}}+P_{{\text{O}}_2}$ (XIII)

Conclusion:

Substitute $\text{4}\text{kg}$ for $m_{\text{He}}$ and $4\text{kg}/\text{kmol}$ for $M_{\text{He}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{He}}=\frac{\text{4}\text{kg}}{4\text{kg}/\text{kmol}}\\ =1\text{kmol}\end{array}$

Substitute $\text{8}\text{kg}$ for $m_{{\text{O}}_2}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_2}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{O}}_2}=\frac{\text{8}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.25\text{kmol}\end{array}$

Substitute $1\text{kmol}$ for $N_{\text{He}}$, and $0.25\text{kmol}$ for $N_{{\text{O}}_2}$ in Equation (III).

$\begin{array}{c}{N}_m=1\text{kmol}+0.25\text{kmol}\\ =1.25\text{kmol}\end{array}$

Substitute $\text{2}\text{.5 kmol}$ for $N_{\text{He}}$, $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R_u$, 280 K for $T_1$, and 250 kPa for $P_{m,1}$ in Equation (IV).

$\begin{array}{c}{\nu }_{\text{He}}=\frac{\left(1\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(170\text{K}\right)}{7\text{MPa}}\\ =\frac{\left(1\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(170\text{K}\right)}{7\text{MPa}\left(\frac{1000\text{kPa}}{1\text{MPa}}\right)}\\ =0.202{\text{m}}^{\text{3}}\end{array}$

Refer Table A-1, “Molar mass, gas constant, and critical2point properties”, obtain the critical temperature and pressure of oxygen as follows.

$\begin{array}{l}{P}_{{\text{cr,O}}_2}=5.08\text{MPa}\\ T_{{\text{cr,O}}_2}=154.8\text{K}\end{array}$

Substitue $7\text{MPa}$ for $P_{m,1}$ and $5.08\text{MPa}$ for $P_{{\text{cr,O}}_2}$ in Equation (V).

$\begin{array}{c}{P}_{R,1}=\frac{7\text{MPa}}{5.08\text{MPa}}\\ =1.38\end{array}$

Substitue $170\text{K}$ for $T_1$ and $154.8\text{K}$ for $T_{{\text{cr,O}}_2}$ in Equation (VI).

$\begin{array}{c}{T}_{R,1}=\frac{170\text{K}}{154.8\text{K}}\\ =1.10\end{array}$

Refer to Figure A-15, obtain the compressibility factor for argon gas by reading the reduced temperature and reduced pressure of $1.10$ and $1.38$.

$Z_{{\text{O}}_2}=0.53$

Substitute $0.53$ for $Z_{{\text{O}}_2}$, $\text{0}\text{.25 kmol}$ for $N_{{\text{O}}_2}$, $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R_u$, $\text{170}\text{K}$ for $T_1$, and $\text{7 MPa}$ for $P_{m,1}$ in Equation (VII).

$\begin{array}{c}{\nu }_{{\text{O}}_2}=\frac{0.53\left(2.5\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(170\text{K}\right)}{\text{7 MPa}}\\ =\frac{0.53\left(2.5\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(170\text{K}\right)}{\text{7 MPa}\left(\frac{1000\text{kPa}}{\text{1}\text{MPa}}\right)}\\ =0.027{\text{m}}^{\text{3}}\end{array}$

Substitute $0.202{\text{m}}^3$ for ${\nu }_{\text{He}}$ and $0.027{\text{m}}^3$ for ${\nu }_{{\text{O}}_2}$ in Equation (VIII).

$\begin{array}{c}{\nu }_{\text{tank}}=0.202{\text{m}}^3+0.027{\text{m}}^3\\ =0.229{\text{m}}^3\end{array}$

Substitute $\text{1}\text{kmol}$ for $N_{{\text{O}}_2}$, $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R_u$, $\text{220}\text{K}$ for $T_1$, and $0.229{\text{m}}^3$ for ${\nu }_{\text{tank}}$ in Equation (IX).

$\begin{array}{c}{P}_{\text{He,2}}=\frac{\left(\text{1}\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(\text{220}\text{K}\right)}{0.229{\text{m}}^3}\\ =7987\text{kPa}\end{array}$

Substitute $220\text{K}$ for $T_2$ and $154.8\text{K}$ for $T_{{\text{cr,O}}_2}$ in Equation (X).

$\begin{array}{c}{T}_{R,{\text{O}}_2}=\frac{220\text{K}}{154.8\text{K}}\\ =1.42\end{array}$

Substitute $0.229{\text{m}}^{\text{3}}$ for ${\nu }_m$, $0.25\text{kmol}$ for $N_{{\text{O}}_2}$, $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$, $154.8\text{K}$ for $T_{\text{cr},{\text{O}}_2}$, and $5080\text{kPa}$ for $P_{\text{cr},{\text{O}}_2}$ in Equation (XI).

$\begin{array}{c}{\nu }_{R,{\text{O}}_2}=\frac{0.229{\text{m}}^{\text{3}}/0.25\text{kmol}}{8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\left(154.8\text{K}\right)/\left(5080\text{kPa}\right)}\\ =3.616\end{array}$

Refer to Figure A-15, obtain the reduced pressure of argon gas by reading the values of reduced temperature and reduced volume of $1.42$ and $3.616$.

$P_{R,{\text{O}}_2}=0.39$

Substitute 0.39 for $P_R$ and $5080\text{kPa}$ for $P_{\text{cr}}$ in Equation (XII).

$\begin{array}{c}{P}_{{\text{O}}_2}=\left(0.39\times 5080\text{kPa}\right)\\ =1981\text{kPa}\left(\frac{1\text{MPa}}{1000\text{kPa}}\right)\\ =1.981\text{MPa}\end{array}$

Substitute $7.987\text{MPa}$ for $P_{\text{He}}$ and $1.981\text{MPa}$ for $P_{{\text{O}}_2}$ in Equation (XIII).

$\begin{array}{c}{P}_{m,2}=7.987\text{MPa}+1.981\text{MPa}\\ \text{= 9}\text{.97}\text{MPa}\end{array}$

Thus, the final pressure of the mixture is $\text{9}\text{.97}\text{MPa}$.

To determine

The amount of heat transfer into the closed system.

Answer to Problem 93RP

The amount of heat transfer into the closed system is $1044.6\text{kJ}$.

Explanation of Solution

Write the expression to obtain the initial and final reduced pressure of ${\text{N}}_2$ $\left(P_{R_1{\text{,N}}_2}\right)$.

$P_{R_1{\text{,N}}_2}=\frac{P_{m,{\text{O}}_2}}{P_{\text{cr},{\text{O}}_2}}$ (XIV)

Write the equation to calculate the reduced temperature of nitrogen gas.

$T_{R,{\text{N}}_{\text{2}}}=\frac{T_m}{T_{\text{cr},{\text{N}}_{\text{2}}}}$ (XV)

Here, mixture temperature is $T_m$.

Write the closed system energy balance relation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$

$\begin{array}{c}{Q}_{\text{in}}=\Delta U\\ =\Delta U_{\text{He}}+\Delta U_{{\text{O}}_2}\\ =m{c}_v\left(T_2-T_1\right)+N_{{\text{O}}_2}\left({\overline{h}}_2-{\overline{h}}_1\right)-\left(P_2{\nu }_2-P_1{\nu }_1\right)\\ =m{c}_v\left(T_2-T_1\right)+N_{{\text{O}}_2}\left({\overline{h}}_2-{\overline{h}}_1\right)-\left(P_{{\text{O}}_2,2}-P_{{\text{O}}_2,1}\right){\nu }_{\text{tank}}\end{array}$ (XVI)

Here, input energy is $E_{\text{in}}$, output energy is $E_{\text{out}}$, change in energy of a system is $\Delta E_{\text{system}}$, change in internal energy of a system is $\Delta U$, and amount of heat transfer into the closed system is $Q_{\text{in}}$.

Write the expression to obtain the initial and final reduced pressure of ${\text{O}}_2$ $\left(P_{R_1{\text{,O}}_2}\right)$.

$P_{R_1{\text{,O}}_2}=\frac{P_{m,{\text{O}}_2}}{P_{\text{cr},{\text{O}}_2}}$ (XVII)

Write formula for enthalpy departure factor $\left(Z_h\right)$ on molar basis.

$Z_h=\frac{{\left({\overline{h}}_{\text{ideal}}-\overline{h}\right)}_{T,P}}{R_u{T}_{\text{cr}}}$ (XVIII)

Here, the molar enthalpy at ideal gas state is ${\overline{h}}_{\text{ideal}}$, the molar enthalpy and normal state is $\overline{h}$, the universal gas constant is $R_u$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (XVIII) to obtain $\overline{h}$.

$\overline{h}={\overline{h}}_{\text{ideal}}-Z_h{R}_u{T}_{\text{cr}}$ (XIX)

Refer Equation (XIX) express as two states of enthalpy difference (final – initial).

${\overline{h}}_2-{\overline{h}}_1={\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}-\left(Z_{h2}-Z_h{}_1\right)R_u{T}_{\text{cr}}$ (XX)

Write the expression to calculate partial pressure at inital stage.

$P_{\text{He,1}}=\frac{N_{\text{He}}{R}_u{T}_1}{{\nu }_{\text{tank}}}$ (XXI)

Write the expression to calculate the total inital pressure.

$P_{{\text{O}}_2}=P_{m,1}-P_{\text{He,1}}$ (XXII)

Conclusion:

Substitute $\text{9}\text{.97}\text{MPa}$ for $P_{m,{\text{O}}_2}$ and $\text{5}\text{.08}\text{MPa}$ for $P_{\text{cr},{\text{O}}_2}$ in Equation (XVII).

$\begin{array}{c}{P}_{R_1{\text{,O}}_2}=\frac{\text{9}\text{.97}\text{MPa}}{\text{5}\text{.08}\text{MPa}}\\ =1.963\end{array}$

Refer Figure A-29, “Generalized enthalpy departure chart”, obtain the value of $Z_{h_1}$ and $Z_{h_2}$ for ${\text{O}}_2$ as $2.2$ and $1.2$ by taking $T_{R_1{\text{,O}}_2}$ as $1.10$ , $P_{R_1{\text{,O}}_2}$ as $1.38$ , $T_{R_2{\text{,O}}_2}$ as $1.2$ and $P_{R_2{\text{,O}}_2}$ as $1.963$.

Substitute $6404\text{kJ/kmol}$ for ${\left({\overline{h}}_2\right)}_{\text{ideal}}$, $4949\text{kJ/kmol}$ for ${\left({\overline{h}}_1\right)}_{\text{ideal}}$$1.2$ for $Z_{h2}$, $2.2$ for $Z_{h1}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $154.8\text{K}$ for $T_{\text{cr}}$ in Equation (XX).

$\begin{array}{c}{\overline{h}}_2-{\overline{h}}_1=\left\{\begin{array}{l}\left(6404\text{kJ/kmol}-4949\text{kJ/kmol}\right)\\ -\left[\left(2.2-1.2\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(154.8\text{K}\right)\right]\end{array}\right\}\\ =2742\text{kJ/kmol}\end{array}$

Substitute $\text{1}\text{kmol}$ for $N_{{\text{O}}_2}$, $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R_u$, $\text{170}\text{K}$ for $T_2$, and $0.229{\text{m}}^3$ for ${\nu }_{\text{tank}}$ in Equation (XXI).

$\begin{array}{c}{P}_{\text{He,1}}=\frac{\left(\text{1}\text{kmol}\right)\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(\text{170}\text{K}\right)}{0.229{\text{m}}^3}\\ =6172\text{kPa}\end{array}$

Substitute $6172\text{kPa}$ for $P_{\text{He,1}}$ and $7\text{MPa}$ for $P_{m,1}$ in Equation (XXII).

$\begin{array}{c}{P}_{{\text{O}}_2,1}=7\text{MPa}-6172\text{kPa}\\ \text{= }7\text{MPa}\left(\frac{1000\text{kPa}}{1\text{MPa}}\right)-6172\text{kPa}\\ =828\text{kPa}\end{array}$

Substitute $4\text{kg}$ for $m$ , $3.1156\text{kJ}/\text{kg}$ for $c_v$ , $220\text{K}$ for $T_2$ , $170\text{K}$ for $T_1$ , $0.25\text{kmol}$ for $N_{{\text{O}}_2}$ , $2742\text{kJ}/\text{kmol}$ for $\left({\overline{h}}_2-{\overline{h}}_1\right)$ , $1981\text{kPa}$ for $P_{{\text{O}}_2,2}$ , $828\text{kPa}$ for $P_{{\text{O}}_2,1}$ and $0.229{\text{m}}^3$ for $v_{\text{tank}}$ in Equation (XVI).

$\begin{array}{c}{Q}_{\text{in}}=\left[\begin{array}{l}\left(4\text{kg}\right)\left(3.1156\text{kJ}/\text{kg}\right)\left(220\text{K}-170\text{K}\right)\\ +0.25\text{kmol}\left(2742\text{kJ}/\text{kmol}\right)-\left(1981\text{kPa}-828\text{kPa}\right)0.229{\text{m}}^3\end{array}\right]\\ =623.1\text{kJ}+421.5\text{kJ}\\ =1044.6\text{kJ}\end{array}$

Thus, the amount of heat transfer into the closed system is $1044.6\text{kJ}$.