Chapter 13.3, Problem 72P

A piston–cylinder device contains 6 kg of H₂ and 21 kg of N₂ at 160 K and 5 MPa. Heat is now transferred to the device, and the mixture expands at constant pressure until the temperature rises to 200 K. Determine the heat transfer during this process by treating the mixture (a) as an ideal gas and (b) as a nonideal gas and using Amagat’s law.

To determine

The heat transfer during the process by treating as an ideal gas.

Answer to Problem 72P

The heat transfer during the process as an ideal gas is $4273\text{kJ}$.

Explanation of Solution

Write a closed system energy balance for the gas mixture.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E^{\nearrow 0\left(\text{steady}\right)}\\ E_{\text{in}}=E_{\text{out}}\\ Q_{\text{in}}-W_{b,out}=\Delta H\end{array}$

$\begin{array}{l}{Q}_{\text{in}}=\Delta H\\ Q_{\text{in}}=\Delta H_{H_2}+\Delta H_{N_2}\\ Q_{\text{in}}=N_{H_2}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{H_2}+N_{{\text{N}}_{\text{2}}}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{N}}_{\text{2}}}\end{array}$ (I)

Here, input energy is $E_{\text{in}}$, output energy is $E_{\text{out}}$, change in energy of a system is $\Delta E_{\text{system}}$, specific amount of heat transfer into the closed system is $Q_{\text{in}}$, and specific enthalpy at state 1 and 2 is ${\overline{h}}_1\text{and}{\overline{h}}_2$ and mole fraction of ${\text{H}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$ is $y_{{\text{O}}_{\text{2}}}\text{and}{y}_{{\text{N}}_{\text{2}}}$ respectively.

Write the expression to obtain the mole number of ${\text{N}}_{\text{2}}$ $\left(N_{{\text{N}}_{\text{2}}}\right)$.

$N_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (II)

Here, molar mass of ${\text{N}}_{\text{2}}$ is $m_{{\text{N}}_{\text{2}}}$ and molar mass of ${\text{N}}_{\text{2}}$ is $M_{{\text{N}}_{\text{2}}}$.

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

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

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of ${\text{N}}_2$ and ${\text{H}}_2$ as $28.0\text{kg}/\text{kmol}$ and $2\text{kg}/\text{kmol}$ respectively.

Substitute $\text{6}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{6}\text{kg}}{2\text{kg}/\text{kmol}}\\ =3\text{kmol}\end{array}$

Substitute $\text{21}\text{kg}$ for $m_{{\text{N}}_2}$ and $28.0\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$ in Equation (III).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{\text{21}\text{kg}}{28.0\text{kg}/\text{kmol}}\\ =0.75\text{kmol}\end{array}$

From the Table of ideal gas for ${\text{H}}_{\text{2}}$, write the value of $h_{1@160\text{K}}$ as $4535.4\text{kJ}/\text{kmol}$ and $h_{2@200\text{K}}$ as $5669.2\text{kJ}/\text{kmol}$, ideal gas for ${\text{N}}_2$, write the value of $h_{1@160\text{K}}$ as $4648\text{kJ}/\text{kmol}$ and $h_{2@200\text{K}}$ as $5810\text{kJ}/\text{kmol}$.

Substitute $3\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $5669.2\text{kJ}/\text{kmol}$ for ${\overline{h}}_{2,{\text{H}}_{\text{2}}}$, $4535.4\text{kJ}/\text{kmol}$ for ${\overline{h}}_{1,{\text{H}}_{\text{2}}}$, $0.75\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $5810\text{kJ}/\text{kmol}$ for ${\overline{h}}_{2,N_{\text{2}}}$, and $4648\text{kJ}/\text{kmol}$ for ${\overline{h}}_{1,N_{\text{2}}}$ to find the heat transfer during the process as an ideal gas in Equation (I).

$\begin{array}{c}{Q}_{\text{in,ideal}}=\left[\begin{array}{l}3\text{kmol}\left(5669.2\text{kJ}/\text{kmol}-4535.4\text{kJ}/\text{kmol}\right)\\ +0.75\text{kmol}\left(5810\text{kJ}/\text{kmol-}4648\text{kJ}/\text{kmol}\right)\end{array}\right]\\ =4273\text{kJ}\end{array}$

Thus, the heat transfer during the process as an ideal gas is $4273\text{kJ}$.

To determine

The heat transfer during the process by treating as non-ideal gas.

Answer to Problem 72P

The heat transfer during the process by treating as non-ideal gas is $4745\text{kJ}$.

Explanation of Solution

Write the expression to obtain the initial reduced temperature of ${\text{H}}_{\text{2}}$ $\left(T_{R1{\text{,H}}_{\text{2}}}\right)$.

$T_{R_1{\text{,H}}_{\text{2}}}=\frac{T_{m,1}}{T_{\text{cr},{\text{H}}_{\text{2}}}}$ (IV)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $T_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the initial and final reduced pressure of ${\text{H}}_{\text{2}}$ $\left(P_{R_1{\text{,H}}_{\text{2}}}\right)$.

$\begin{array}{c}{P}_{R_1{\text{,H}}_{\text{2}}}=P_{R_2,H_2}\\ =\frac{P_m}{P_{\text{cr},{\text{H}}_{\text{2}}}}\end{array}$ (V)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $P_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the final reduced temperature of ${\text{H}}_{\text{2}}$ $\left(T_{R_2{\text{,H}}_{\text{2}}}\right)$.

$T_{R_2{\text{,H}}_{\text{2}}}=\frac{T_{m,2}}{T_{\text{cr},{\text{H}}_{\text{2}}}}$ (VI)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $T_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the initial reduced temperature of ${\text{N}}_2$ $\left(T_{R1{\text{,N}}_2}\right)$.

$T_{R_1{\text{,N}}_2}=\frac{T_{m,1}}{T_{\text{cr},{\text{N}}_2}}$ (VII)

Here, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$.

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

$\begin{array}{c}{P}_{R_1{\text{,N}}_2}=P_{R_2,{\text{N}}_2}\\ =\frac{P_m}{P_{\text{cr},{\text{N}}_2}}\end{array}$ (VIII)

Here, critical temperature of ${\text{N}}_2$ is $P_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the final reduced temperature of ${\text{N}}_2$ $\left(T_{R_2{\text{,N}}_2}\right)$.

$T_{R_2{\text{,N}}_2}=\frac{T_{m,2}}{T_{\text{cr},{\text{N}}_2}}$ (IX)

Here, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$.

Consider hydrogen as ideal gas

Write the expression for molar enthalpy difference of hydrogen ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{H}}_{\text{2}}}$.

${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{H}}_{\text{2}}}={\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{H}}_{\text{2}}\text{,ideal}}$ (X)

Write the expression for molar enthalpy difference of nitrogen.

${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{N}}_{\text{2}}}=R_u{T}_{\text{cr}}\left(Z_{h2}-Z_h{}_1\right)+{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}$ (XI)

Conclusion:

Substitute 160 K for $T_{m,1}$ and 33.3 K for $T_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{c}{T}_{R_1{\text{,H}}_{\text{2}}}=\frac{160\text{}\text{K}}{33.3\text{K}}\\ =4.805\end{array}$

Substitute 5 MPa for $P_m$ and 1.30 MPa for $P_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (V).

$\begin{array}{c}{P}_{R{\text{,H}}_{\text{2}}}=\frac{5\text{MPa}}{1.30\text{MPa}}\\ =3.846\end{array}$

Substitute 200 K for $T_{m,2}$ and 33.3 K for $T_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (VI).

$\begin{array}{c}{T}_{R_2{\text{,H}}_{\text{2}}}=\frac{200\text{}\text{K}}{33.3\text{K}}\\ =6.006\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of $Z_{h_1}$ and $Z_{h_2}$ for ${\text{H}}_{\text{2}}$ as 0 and 0 by taking $T_{R_1{\text{,H}}_{\text{2}}}$ as 4.805, $P_{R_1{\text{,H}}_{\text{2}}}$ as 3.846, and $T_{R_2{\text{,H}}_{\text{2}}}$ as 6.006.

Substitute 160 K for $T_{m,1}$ and 126.2 K for $T_{\text{cr},{\text{N}}_2}$ in Equation (VII).

$\begin{array}{c}{T}_{R_1{\text{,N}}_2}=\frac{160\text{}\text{K}}{126.2\text{K}}\\ =1.268\end{array}$

Substitute 5 MPa for $P_m$ and 3.39 MPa for $P_{\text{cr},{\text{N}}_2}$ in Equation (VIII).

$\begin{array}{c}{P}_{R{\text{,N}}_2}=\frac{5\text{MPa}}{3.39\text{MPa}}\\ =1.475\end{array}$

Substitute 200 K for $T_{m,2}$ and 126.2 K for $T_{\text{cr},{\text{N}}_2}$ in Equation (IX).

$\begin{array}{c}{T}_{R_2{\text{,N}}_2}=\frac{200\text{}\text{K}}{126.2\text{K}}\\ =1.585\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of $Z_{s_1}$ and $Z_{s_2}$ for ${\text{N}}_2$ as 0.8 and 0.4 by taking $T_{R_1{\text{,N}}_2}$ as 1.268, $P_{R_1{\text{,N}}_2}$ as 1.475 and $T_{R_2{\text{,N}}_2}$ as 1.585.

Substitute $5669.2\text{kJ}/\text{kmol}$ for ${\overline{h}}_2$ and $4535.4\text{kJ}/\text{kmol}$ for ${\overline{h}}_1$ in Equation (X).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{H}}_{\text{2}}}=\left(5669.2\text{kJ}/\text{kmol}-4535.4\text{kJ}/\text{kmol}\right)\\ =1133.8\text{kJ}/\text{kmol}\end{array}$

Substitute $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$ , $126.2\text{K}$ for $T_{cr}$ , $1.3$ for $Z_{h2}$, $0.7$ for $Z_h{}_1$, $5810\text{kJ}/\text{mol}$ for ${\overline{h}}_2$ and $4648\text{kJ}/\text{mol}$ for ${\overline{h}}_1$ in Equation (XI).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{N}}_{\text{2}}}=\left[\begin{array}{l}\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\left(126.2\text{K}\right)\left(1.3-0.7\right)\\ +\left(5810\text{kJ}/\text{mol}-4648\text{kJ}/\text{mol}\right)\end{array}\right]\\ =1791.5\text{kJ}/\text{mol}\end{array}$

Substitute $3\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $5669.2\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{N}}_{\text{2}}}$, $4535.4\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{{\text{H}}_{\text{2}}}$ and $0.75\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{Q}_{\text{in}}=\left[3\text{kmol}\left(1133.8\text{kJ}/\text{kmol}\right)+0.75\text{kmol}\left(1791.5\text{kJ}/\text{kmol}\right)\right]\\ =4745\text{kJ}\end{array}$

Thus, the heat transfer during the process by treating as non-ideal gas is $4745\text{kJ}$.