Chapter 18, Problem 61P

(a)

To determine

Pressure and volume at given points.

Answer to Problem 61P

Volume at A, B, C, D are $\underset{\_}{1.97\times {10}^{-3}{\text{m}}^{\text{3}}}$, $\underset{\_}{11.9\times {10}^{-3}{\text{m}}^{\text{3}}}$, $\underset{\_}{32.8\times {10}^{-3}{\text{m}}^{\text{3}}}$ and $\underset{\_}{5.44\times {10}^3{\text{m}}^{\text{3}}}$ respectively

Pressure at B and D are $\underset{\_}{4.14\text{atm}}$ and $\underset{\_}{6.03\text{atm}}$ respectively.

Explanation of Solution

Write the ideal gas equation and rearrange for $V_A$ and $V_C$ to determine the volume at point A and C.

    $\begin{array}{c}{P}_A{V}_A=nR{T}_A\\ V_A=\frac{nR{T}_A}{P_A}\\ P_A{V}_C=nR{T}_C\\ V_C=\frac{nR{T}_C}{P_C}\end{array}$        (I)

Here $P_A$, and $P_C$ is the pressure at A and C respectively, $R$ is the real gas constant, $n$ is the number of moles, $V_A$ and $V_C$ is the volume at A and C respectively, $T_A$ and $T_C$ is the temperature at A and C respectively.

From figure, AB is isothermal.

Therefore.

    $P_A{V}_A=P_B{V}_B$        (II)

Here $P_B$ is the pressure at B and $V_B$ is the volume at B.

BC is adiabatic.

Therefore,

    $P_C{V}_C{}^{\gamma }=P_B{V}_B{}^{\gamma }$        (III)

Here $\gamma$ is the specific heat ratio.

Rearrange (II) in terms of $P_B$ and substitute in (III)

    $\left(\frac{P_A{V}_A}{V_B}\right)V_B{}^{\gamma }=P_C{V}_C{}^{\gamma }$        (IV)

Rearrange (IV) in terms of $V_B$

    $V_B=\left(\frac{P_C}{P_A}\right){\left(\frac{V_C{}^{\gamma }}{V_A}\right)}^{\frac{1}{{}^{\gamma -1}}}$        (V)

Similarly,

CD is isothermal.

    $P_C{V}_C=P_D{V}_D$        (VI)

Here $P_D$ is the pressure at D and $V_D$ is the volume at D.

AD is adiabatic.

    $P_A{V}_A{}^{\gamma }=P_D{V}_D{}^{\gamma }$        (VII)

Substitute for $P_D$ from (VI) in (VII)

    $P_A{V}_A{}^{\gamma }=\left(\frac{P_C{V}_C}{V_D}\right)V_D{}^{\gamma }$        (VIII)

Rearrange (VIII) in terms of $V_D$

    $V_D=\left(\frac{P_A}{P_C}\right){\left(\frac{V_A{}^{\gamma }}{V_C}\right)}^{\frac{1}{{}^{\gamma -1}}}$        (IX)

Rearrange (II) and (VI) in terms of $P_B$ and $P_D$

    $\begin{array}{l}{P}_B=\frac{P_A{V}_A}{V_B}\\ P_D=\frac{P_C{V}_C}{V_D}\end{array}$        (X)

Conclusion:

Substitute $1\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $600\text{K}$ for $T_A$, $400\text{K}$ for $T_C$, $25\text{atm}$ for $P_A$ and $1\text{atm}$ for $P_C$ in (I)

    $\begin{array}{c}{V}_A=\frac{\left(1\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)}{25\text{atm}\left(1.013\times {10}^5\text{Pa}\right)}\\ =1.97\times {10}^{-3}\text{m}\end{array}$

    $\begin{array}{c}{V}_C=\frac{\left(1\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(400\text{K}\right)}{1\text{atm}\left(1.013\times {10}^5\text{Pa}\right)}\\ =32.8\times {10}^{-3}\text{m}\end{array}$

Substitute $1\text{atm}$ for $P_C$, $25\text{atm}$ for $P_A$, $32.8\times {10}^{-3}\text{m}$ for $V_C$, $1.97\times {10}^{-3}\text{m}$ for $V_A$ and $1.4$ for $\gamma$ in (V)

    $\begin{array}{c}{V}_B=\left(\frac{1\text{atm}}{25\text{atm}}\right){\left(\frac{{\left(32.8\times {10}^{-3}\text{m}\right)}^{1.4}}{1.97\times {10}^{-3}\text{m}}\right)}^{\frac{1}{{}^{1.4-1}}}\\ =11.9\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Substitute $1\text{atm}$ for $P_C$, $25\text{atm}$ for $P_A$, $32.8\times {10}^{-3}\text{m}$ for $V_C$, $1.97\times {10}^{-3}\text{m}$ for $V_A$ and $1.4$ for $\gamma$ in (IX)

    $\begin{array}{c}{V}_D=\left(\frac{25\text{atm}}{1\text{atm}}\right){\left(\frac{{\left(1.97\times {10}^{-3}\text{m}\right)}^{1.4}}{32.8\times {10}^{-3}\text{m}}\right)}^{\frac{1}{{}^{1.4-1}}}\\ =5.44\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Substitute $1.97\times {10}^{-3}\text{m}$ for $V_A$, $11.9\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_B$, $32.8\times {10}^{-3}\text{m}$ for $V_C$, $5.44\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_D$, $1\text{atm}$ for $P_C$ and $25\text{atm}$ for $P_A$ in (X)

    $\begin{array}{c}{P}_B=\frac{\left(1.97\times {10}^{-3}\text{m}\right)\left(25\text{atm}\right)}{\left(11.9\times {10}^{-3}{\text{m}}^{\text{3}}\right)}\\ =4.14\text{atm}\\ P_D=\frac{\left(32.8\times {10}^{-3}\text{m}\right)\left(1\text{atm}\right)}{\left(5.44\times {10}^{-3}{\text{m}}^{\text{3}}\right)}\\ =6.03\text{atm}\end{array}$

Volume at A, B, C, D are $\underset{\_}{1.97\times {10}^{-3}{\text{m}}^{\text{3}}}$, $\underset{\_}{11.9\times {10}^{-3}{\text{m}}^{\text{3}}}$, $\underset{\_}{32.8\times {10}^{-3}{\text{m}}^{\text{3}}}$ and $\underset{\_}{5.44\times {10}^3{\text{m}}^{\text{3}}}$ respectively

Pressure at B and D are $\underset{\_}{4.14\text{atm}}$ and $\underset{\_}{6.03\text{atm}}$ respectively.

(b)

To determine

Net work done per cycle.

Answer to Problem 61P

Net work done is $\underset{\_}{2.99\text{kJ}}$.

Explanation of Solution

Write the first law of thermodynamics

    $\Delta E=Q_h+W_{AB}$        (XI)

Here $\Delta E$ is the change in internal energy, $Q_h$ is the amount of heat transfer and $W_{AB}$ is the work done.

As the change in internal energy is zero, (XI) becomes,

    $Q_h=-W_{AB}$

Write the equation for work done

    $\left|Q_h\right|=nR{T}_h\ln \left(\frac{V_B}{V_A}\right)$        (XII)

Here $T_h$ is the temperature of the hot reservoir.

Write the equation for efficiency

    $e=\frac{W_{eng}}{\left|Q_h\right|}$        (XIII)

Here $e$ is the efficiency of the engine and $W_{eng}$ is the work done by the engine.

Rewrite (XIIII) in terms of $W_{eng}$ and substitute for $Q_h$ from (XII)

    $W_{\text{eng}}=enR{T}_h\ln \left(\frac{V_B}{V_A}\right)$        (XIV)

Conclusion:

Substitute $1.97\times {10}^{-3}\text{m}$ for $V_A$, $11.9\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_B$, $\text{1}\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $600\text{K}$ for $T_h$ and $0.33$ for $e$ in (XIV)

    $\begin{array}{c}{W}_{\text{eng}}=\left(0.33\right)\left(\text{1}\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)\ln \left(\frac{11.9\times {10}^{-3}{\text{m}}^{\text{3}}}{1.97\times {10}^{-3}\text{m}}\right)\\ =2.99\text{kJ}\end{array}$

Net work done is $\underset{\_}{2.99\text{kJ}}$.