Chapter 13.3, Problem 94RP

A spring-loaded piston–cylinder device contains a mixture of gases whose pressure fractions are 25 percent Ne, 50 percent O₂, and 25 percent N₂. The piston diameter and spring are selected for this device such that the volume is 0.1 m³ when the pressure is 200 kPa and 1.0 m³ when the pressure is 1000 kPa. Initially, the gas is added to this device until the pressure is 200 kPa and the temperature is 10°C. The device is now heated until the pressure is 500 kPa. Calculate the total work and heat transfer for this process.

To determine

The total work done and heat transfer for the process.

Answer to Problem 94RP

The total work done for the process is $118\text{kJ}$.

The heat transfer for the process is $569\text{kJ}$.

Explanation of Solution

Write the expression to calculate the partial pressure of $\text{Ne}$ $\left(\text{Ne}\right)$.

$P_{\text{Ne}}=y_{\text{Ne}}{P}_m$ (I)

Here, mixture pressure is $P_m$ and mole fraction of $\text{Ne}$ is $y_{\text{Ne}}$.

Write the expression to calculate the partial pressure of ${\text{O}}_2$ $\left(P_{{\text{O}}_2}\right)$.

$P_{{\text{O}}_2}=y_{{\text{O}}_2}{P}_m$ (II)

Here, mixture pressure is $P_m$ and mole fraction of ${\text{O}}_2$ is $y_{{\text{O}}_2}$.

Write the expression to calculate the partial pressure of ${\text{N}}_{\text{2}}$ $\left(P_{{\text{N}}_{\text{2}}}\right)$.

$P_{{\text{N}}_{\text{2}}}=y_{{\text{N}}_{\text{2}}}{P}_m$ (III)

Here, mixture pressure is $P_m$ and mole fraction of ${\text{N}}_2$ is $y_{{\text{N}}_2}$.

Write the expression to calculate the mass of ${\text{N}}_{\text{2}}$ $\left(m_{{\text{N}}_{\text{2}}}\right)$.

$m_{\text{Ne}}=\frac{P_{\text{Ne}}{\nu }_{\text{m}}}{R_{\text{Ne}}T}$ (IV)

Write the expression to calculate the mass of ${\text{O}}_{\text{2}}$ $\left(m_{{\text{O}}_{\text{2}}}\right)$.

$m_{{\text{O}}_{\text{2}}}=\frac{P_{{\text{O}}_{\text{2}}}{\nu }_{\text{m}}}{R_{{\text{O}}_{\text{2}}}T}$ (V)

Write the expression to calculate the mass of ${\text{N}}_{\text{2}}$ $\left(m_{{\text{N}}_{\text{2}}}\right)$.

$m_{{\text{N}}_{\text{2}}}=\frac{P_{{\text{N}}_{\text{2}}}{\nu }_{\text{m}}}{R_{{\text{N}}_{\text{2}}}T}$ (VI)

Write the expression to calculate the total mass of each component $\left(m_{\text{total}}\right)$.

$m_{\text{total}}=m_{\text{Ne}}+m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}$ (VII)

Write the expression to calculate the mass fraction of $\text{Ne}$$\left({\text{mf}}_{\text{Ne}}\right)$

${\text{mf}}_{\text{Ne}}=\frac{m_{\text{Ne}}}{m_m}$ (VIII)

Write the expression to calculate the mass fraction of ${\text{O}}_{\text{2}}$$\left({\text{mf}}_{{\text{O}}_{\text{2}}}\right)$

${\text{mf}}_{{\text{O}}_{\text{2}}}=\frac{m_{{\text{O}}_{\text{2}}}}{m_m}$ (IX)

Write the expression to calculate the mass fraction of ${\text{N}}_{\text{2}}$$\left({\text{mf}}_{{\text{N}}_{\text{2}}}\right)$

${\text{mf}}_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{m_m}$ (X)

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

$N_{\text{Ne}}=\frac{m_{\text{Ne}}}{M_{\text{Ne}}}$ (XI)

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

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

$N_{{\text{N}}_2}=\frac{m_{{\text{N}}_2}}{M_{{\text{N}}_2}}$ (XII)

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

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

$N_{{\text{CO}}_2}=\frac{m_{{\text{CO}}_2}}{M_{{\text{CO}}_2}}$ (XIII)

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

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

$N_m=N_{{\text{CO}}_2}+N_{{\text{N}}_2}$ (XIV)

Write the expression to calculate the apparent molecular weight of the mixture $\left(M_m\right)$.

$M_m=\frac{m_m}{N_m}$ (XV)

Write the expression to calculate the constant volume specific heat of the mixture $\left(c_v\right)$.

$c_v={\text{mf}}_{\text{Ne}}{c}_v{}_{\text{Ne}}+{\text{mf}}_{{\text{O}}_2}{c}_v{}_{,{\text{O}}_2}{\text{+mf}}_{{\text{N}}_2}{c}_v{}_{{\text{N}}_2}$ (XVI)

Here, mole fraction of ${\text{Ne,N}}_{\text{2}}\text{,}\text{and}{\text{O}}_{\text{2}}$ is ${\text{mf}}_{\text{Ne}},{\text{mf}}_{{\text{N}}_2},\text{and}{\text{mf}}_{{\text{O}}_2}$, specific heat of $\text{Ne}$ is $c_{v,}{}_{\text{Ne}}$, specific heat of ${\text{N}}_{\text{2}}$ is $c_{v,}{}_{{\text{N}}_{\text{2}}}$, and specific heat of ${\text{O}}_{\text{2}}$ is $c_{v,}{}_{{\text{O}}_{\text{2}}}$.

Write the expression to calculate the apparent gas constant of the mixture $\left(R\right)$.

$R=\frac{R_u}{M_m}$ (XVII)

Here, universal gas constant is $R_u$.

Write the expression for the mass contained in the system $\left(m\right)$.

$m=\frac{P_1{\nu }_1}{R{T}_1}$ (XVIII)

Write the expression to calculate the final temperature $\left(T_2\right)$

$T_2=\frac{P_2{\nu }_2}{mR}$ (XIX)

Write the expression to calculate the work done during the process.

$W_{out}=\frac{P_1+P_2}{2}\left({\nu }_2-{\nu }_1\right)$ (XX)

Conclusion:

From Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the values of molar masses for $\text{Ne}$,${\text{O}}_2$ and ${\text{N}}_2$ as $20.18\text{kg}/\text{kmol},32.0\text{kg}/\text{kmol}$ and $28.0\text{kg}/\text{kmol}$ respectively.

From Table A-2a, “Ideal-gas specific heats of various common gases”, obtain the following properties for $\text{Ne}$,${\text{O}}_2$ and ${\text{N}}_2$.

For $\text{Ne}$,

$c_v=0.6179\text{kJ}/\text{kg}\cdot \text{K}$

For ${\text{O}}_2$,

$c_v=0.658\text{kJ}/\text{kg}\cdot \text{K}$

For ${\text{N}}_2$,

$c_v=0.743\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $0.25$ 0for $y_{\text{Ne}}$ and $200\text{kPa}$ for $P_m$ in Equation (I).

$\begin{array}{c}{P}_{{\text{O}}_{\text{2}}}=\left(0.25\right)200\text{kPa}\\ =50\text{kPa}\end{array}$

Substitute $0.50$ for $y_{{\text{O}}_2}$ and $200\text{kPa}$ for $P_m$ in Equation (II).

$\begin{array}{c}{P}_{{\text{O}}_2}=\left(0.50\right)200\text{kPa}\\ =100\text{kPa}\end{array}$

Substitute $0.25$ for $y_{{\text{N}}_{\text{2}}}$ and $200\text{kPa}$ for $P_m$ in Equation (III).

$\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\left(0.25\right)200\text{kPa}\\ =50\text{kPa}\end{array}$

Substitute $50\text{kPa}$ for $P_{\text{Ne}}$, $0.1{\text{m}}^3$ for ${\nu }_{\text{m}}$, $0.4119\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{\text{Ne}}$ and $283\text{K}$ in Equation (IV).

$\begin{array}{c}{m}_{\text{Ne}}=\frac{\left(50\text{kPa}\right)\times \left(0.1{\text{m}}^3\right)}{\left(0.4119\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(283\text{K}\right)}\\ =0.4289\text{kg}\end{array}$

Substitute $100\text{kPa}$ for $P_{{\text{O}}_{\text{2}}}$, $0.1{\text{m}}^3$ for ${\nu }_{\text{m}}$, $0.2598\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{{\text{O}}_{\text{2}}}$ and $283\text{K}$ in Equation (V).

$\begin{array}{c}{m}_{{\text{O}}_{\text{2}}}=\frac{\left(100\text{kPa}\right)\times \left(0.1{\text{m}}^3\right)}{\left(0.2598\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(283\text{K}\right)}\\ =0.1360\text{kg}\end{array}$

Substitute $50\text{kPa}$ for $P_{{\text{N}}_{\text{2}}}$, $0.1{\text{m}}^3$ for ${\nu }_{\text{m}}$, $0.2968\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{{\text{N}}_{\text{2}}}$ and $283\text{K}$ in Equation (VI).

$\begin{array}{c}{m}_{{\text{N}}_{\text{2}}}=\frac{\left(50\text{kPa}\right)\times \left(0.1{\text{m}}^3\right)}{\left(0.2968\text{kpa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(283\text{K}\right)}\\ =0.05953\text{kg}\end{array}$

Substitute $0.4289\text{kg}$ for $m_{\text{Ne}}$, $0.1360\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$, and $0.05953\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ in Equation (VII).

$\begin{array}{c}{m}_{\text{total}}=0.4289\text{kg}+0.1360\text{kg}+0.05953\text{kg}\\ =0.2384\text{kg}\end{array}$

Substitute $0.04289\text{kg}$ for $m_{\text{Ne}}$, and $0.2384\text{kg}$ for $m_m$ in Equation (VIII).

$\begin{array}{c}{\text{mf}}_{\text{Ne}}=\frac{0.04289\text{kg}}{0.2384\text{kg}}\\ =0.1799\end{array}$

Substitute $0.1360\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$, and $0.2384\text{kg}$ for $m_m$ in Equation (IX).

$\begin{array}{c}{\text{mf}}_{{\text{O}}_{\text{2}}}=\frac{0.1360\text{kg}}{0.2384\text{kg}}\\ =0.5705\end{array}$

Substitute $0.05953\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$, and $0.2384\text{kg}$ for $m_m$ in Equation (X).

$\begin{array}{c}{\text{mf}}_{{\text{N}}_{\text{2}}}=\frac{0.05953\text{kg}}{0.2384\text{kg}}\\ =0.2497\end{array}$

Substitute $0.04289\text{kg}$ for $m_{\text{Ne}}$ and $20.18\text{kg}/\text{mol}$ for $M_{\text{Ne}}$ in Equation (XI).

$\begin{array}{c}{N}_{\text{Ne}}=\frac{0.04289\text{kg}}{20.18\text{kg}/\text{mol}}\\ =0.002126\text{kmol}\end{array}$

Substitute $0.05953\text{kg}$ for $m_{{\text{N}}_2}$ and $28\text{kg}/\text{mol}$ for $M_{{\text{N}}_2}$ in Equation (XII).

$\begin{array}{c}{N}_{{\text{N}}_2}=\frac{0.05953\text{kg}}{28\text{kg}/\text{mol}}\\ =0.002126\text{kmol}\end{array}$

Substitute $45\text{kg}$ for $m_{{\text{CO}}_2}$ and $28\text{kg}/\text{mol}$ for $M_{{\text{CO}}_2}$ in Equation (XIII).

$\begin{array}{c}{N}_{{\text{O}}_2}=\frac{0.1360\text{kg}}{32\text{kg}/\text{mol}}\\ =0.00425\text{kmol}\end{array}$

Substitute $0.002126\text{kmol}$ for $N_{\text{Ne}}$, $0.00425\text{kmol}$ for $N_{{\text{O}}_2}$, and $0.002126\text{kmol}$ for $N_{{\text{N}}_2}$ in Equation (XIV).

$\begin{array}{c}{N}_m=0.002126\text{kmol}+0.00425\text{kmol}+0.002126\text{kmol}\\ =0.008502\text{kmol}\end{array}$

Substitute $0.008502\text{kmol}$ for $N_m$ and $2.987\text{kmol}$ for $m_m$ in Equation (XV).

$\begin{array}{c}{M}_m=\frac{100\text{kg}}{0.008502\text{kmol}}\\ =28.04\text{kg}/\text{kmol}\end{array}$

Substitute $0.6179\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v{}_{\text{Ne}}$, $0.743\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{v,}{}_{{\text{N}}_{\text{2}}}$, $0.658\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{v,}{}_{{\text{O}}_{\text{2}}}$, and $0.1799$ for ${\text{mf}}_{\text{Ne}}$ , $0.5705$ for ${\text{mf}}_{{\text{O}}_2}$ and $0.2497$ for ${\text{mf}}_{{\text{N}}_2}$ in Equation (XVI).

$\begin{array}{c}{c}_v=\left(0.1799\right)\left(0.6179\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.5705\right)\left(0.658\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.2497\right)\left(0.743\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.672\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $8.134\text{kJ}/\text{Kmol}\cdot \text{K}$ for $R_u$ and $28.04\text{kg}/\text{kmol}$ for $M_m$ in Equation (XVII).

$\begin{array}{c}R=\frac{8.134\text{kJ}/\text{Kmol}\cdot \text{K}}{28.04\text{kg}/\text{kmol}}\\ =0.2964\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $200\text{kPa}$ for $P_1$, $0.1{\text{m}}^3$ for ${\nu }_1$ , $0.2964\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for R and $283\text{K}$ for $T_1$ in Equation (XVIII).

$\begin{array}{c}m=\frac{200\text{kPa}\times 0.1{\text{m}}^3}{0.2964\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\times 283\text{K}}\\ =0.2384\text{kg}\end{array}$

The pressure changes linearly with volume as shown is Figure (1).

Using the data form Prob. 13–94 obtain the value of final volume by linear interpolation.

Write the straight line equation for two points.

$\frac{\left(y_1-y\right)}{\left(y_2-y\right)}=\frac{\left(x_1-x\right)}{\left(x_2-x\right)}$ (XXI)

Here, coordinates of the point 1 is $\left(x_1,y_1\right)$, coordinates of the point 2 is $\left(x_2,y_2\right)$ and intermediate point between 1 and 2 is $\left(x,y\right)$.

Substitute $500$ for $y_1$ , $200$ for $y$ , $1000$ for $y_2$ , $0.1$ for $x$ , $1.0$ for $x_2$ and $0.1$ for $x_2$ in Equation (XXI).

$\begin{array}{l}\frac{\left(500-200\right)}{\left(1000-200\right)}=\frac{\left(x_1-0.1\right)}{\left(1.0-0.1\right)}\\ x_1=0.4375\end{array}$

The final volume ${\nu }_2$ is $0.4375{\text{m}}^3$.

Substitute $500\text{kPa}$ for $P_1$, $0.4375{\text{m}}^3$ for ${\nu }_2$ , $0.2964\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for R and $0.2384\text{kg}$ for $m$ in Equation (XIX).

$\begin{array}{c}{T}_2=\frac{500\text{kPa}\times 0.4375{\text{m}}^3}{0.2384\text{kg}\times 0.2483\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}}\\ =3096\text{K}\end{array}$

Substitute $500\text{kPa}$ for $P_1$ , $200\text{kPa}$ , $0.1{\text{m}}^3$ for ${\nu }_1$ and $0.4375{\text{m}}^3$ for ${\nu }_2$ in Equation (XX).

$\begin{array}{c}{W}_{out}=\frac{500\text{kPa}+200\text{kPa}}{2}\left(0.4375{\text{m}}^3-0.1{\text{m}}^3\right)\\ =118\text{kPa}\cdot {\text{m}}^3\left(\frac{\text{1}\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^3}\right)\\ =118\text{kJ}\end{array}$

Thus, the total work done for the process is $118\text{kJ}$.

Write a energy balance on the system.

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

Here, input energy transfer and output energy transfer is $E_{\text{in}}\text{and}{E}_{\text{out}}$, and change in energy of a system is $\Delta E_{\text{system}}$.

The rate of change in energy of a system $\Delta {\dot{E}}_{\text{system}}$ is zero for steady state.

For given system the energy balance Equation (XXII) is expressed as follows:

$Q_{in}=W_{out}+m{c}_v\left(T_2-T_1\right)$ (XXIII).

The rate of change in energy of a system $\Delta {\dot{E}}_{\text{system}}$ is zero for steady state.

Substitute $118\text{kJ}$ for $W_{out}$, $0.2384\text{kg}$ for $m$ , $0.672\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$ , $3096\text{K}$ for $T_2$ and $283\text{K}$ for $T_1$ in Equation (XXIII).

$\begin{array}{c}{Q}_{in}=118\text{kJ}+\left(0.2384\text{kg}\right)\left(0.672\text{kJ}/\text{kg}\cdot \text{K}\right)\left(3096\text{K}-283\text{K}\right)\\ =569\text{kJ}\end{array}$

Thus, the heat transfer for the process is $569\text{kJ}$.