Chapter 22, Problem 73PQ

Figure P22.73 illustrates the cycle ABCA for a 2.00-mol sample of an ideal diatomic gas, where the process CA is a reversible isothermal expansion. What is a. the net work done by the gas during one cycle? b. How much energy is added to the gas by heat during one cycle? c. How much energy is exhausted from the gas by heat during one cycle? d. What is the efficiency of the cycle? e. What would be the efficiency of a Carnot engine operated between the temperatures at points A and B during each cycle?

To determine

The net work done by the gas during one cycle.

Answer to Problem 73PQ

The net work done by the gas during one cycle is $-411\text{J}$.

Explanation of Solution

The work done by the gas along AB is $W_{AB}$, has magnitude zero since the process is isochoric process.

Write the expression to calculate the work done by the gas along BC.

  $W_{BC}=P\left(V_f-V_i\right)$                                                                                                      (I)

Here, $W_{BC}$ is the work done along BC, P is the pressure of the gas, $V_f$ is the final volume and $V_i$ is the initial volume.

Write the expression to calculate the work done by the gas along CA.

  $W_{CA}=P{V}_i\ln \left(\frac{V_f}{V_i}\right)$                                                                                                   (II)

Here, $W_{CA}$ is the work done by the gas along CA, $V_i$ is the initial volume and $V_f$ is the final volume.

Write the expression to calculate the total work done by the gas.

  $W=W_{AB}+W_{BC}+W_{CA}$                                                                                              (III)

Here, W is the total work done by the gas.

Conclusion:

Substitute $4.50\text{atm}$ for P, $3.00\text{L}$ for $V_i$ and $1.00\text{L}$ for $V_f$ in the above equation (I) to calculate $W_{BC}$.

  $\begin{array}{c}{W}_{BC}=\left(4.50\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right)\left(3.00\text{L}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)-1.00\text{L}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)\\ =-912\text{J}\end{array}$

Substitute $4.50\text{atm}$ for P, $1.00\text{L}$ for $V_i$ and $3.00\text{L}$ for $V_f$ in the above equation (II) to calculate $W_{BC}$.

  $\begin{array}{c}{W}_{CA}=\left(4.50\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right)1.00\text{L}{\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)}_i\ln \left(\frac{3.00\text{L}}{1\text{L}}\right)\\ =501\text{J}\end{array}$

Substitute $0$ for $W_{AB}$, $-912\text{J}$ for $W_{BC}$ and $501\text{J}$ for $W_{CA}$ in the above equation (IV) to calculate W.

  $\begin{array}{c}W=0+\left(-912\text{J}\right)+501\text{J}\\ =-411\text{J}\end{array}$

Therefore, the net work done by the gas during one cycle is $-411\text{J}$.

To determine

The heat added in one cycle.

Answer to Problem 73PQ

The heat added in one cycle is $2780\text{J}$.

Explanation of Solution

For the diatomic gases, the specific heat capacity at constant volume is $\frac{5}{2}R$ and specific heat capacity at constant pressure is $\frac{7}{2}R$. The heat absorbed during the isothermal process along CA is same as the work done the gas along the same path.

The temperature at A and C of the gas is same.

Write the expression to calculate the temperature of the gas at A and C.

  $T=\frac{PV}{nR}$                                                                                                                     (V)

Here, T is the temperature of the gas at A and C, P is the pressure at A, V is the pressure at A n is the number of moles and R is the universal gas constant.

Write the expression to calculate the temperature at B.

  $T^{\prime }=\frac{P^{\prime }{V}^{\prime }}{nR}$                                                                                                                (VI)

Here, $T^{\prime }$ is he temperature at B, $P^{\prime }$ is the pressure at B and $V^{\prime }$ is the volume at B.

Write the expression to calculate the energy added to the system.

  $Q=n{c}_v\left(T^{\prime }-T\right)+W_{CA}$

Here, Q is the energy added to the system, $c_v$ is the specific heat capacity at constant volume and $c_p$ is the specific heat capacity at constant pressure.

Substitute $\frac{5}{2}R$ for $c_v$ in the above equation to rewrite.

  $Q=n\left(\frac{5}{2}R\right)\left(T^{\prime }-T\right)+W_{CA}$                                                                     (VII)

Conclusion:

Substitute $1.50\text{atm}$ for $P^{\prime }$, $3.00\text{L}$ for V, $2.00\text{mol}$ for n and $8.314\text{J}/\text{mol}\cdot \text{K}$ for in the equation (V) to calculate T.

    $\begin{array}{c}T=\frac{\left(1.50\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right)3.00\text{L}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)}{\left(2.00\text{mol}\right)8.314\text{J}/\text{mol}\cdot \text{K}}\\ =27.4\text{K}\end{array}$

Substitute $4.50\text{atm}$ for P, $3.00\text{L}$ for V, $2.00\text{mol}$ for n and $8.314\text{J}/\text{mol}\cdot \text{K}$ for R in the equation (VI) to calculate T.

    $\begin{array}{c}{T}^{\prime }=\frac{\left(4.50\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right)3.00\text{L}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)}{\left(2.00\text{mol}\right)8.314\text{J}/\text{mol}\cdot \text{K}}\\ =82.2\text{K}\end{array}$

Substitute $2.00\text{mol}$ for n and $8.314\text{J}/\text{mol}\cdot \text{K}$ for R, $27.4\text{K}$ for T, $82.2\text{K}$ for $T^{\prime }$ and $501\text{J}$ for $W_{CA}$ in the above equation (VII) to calculate Q.

    $\begin{array}{c}Q=\left(2.00\text{mol}\right)\left(\frac{5}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(82.2\text{K}-27.4\text{K}\right)+501\text{J}\\ \approx 2780\text{J}\end{array}$

Therefore, the heat added in one cycle is $2780\text{J}$.

To determine

The energy exhausted from the gas.

Answer to Problem 73PQ

The energy exhausted from the gas is $3.19\times {10}^3\text{J}$.

Explanation of Solution

For the diatomic gases, the specific heat capacity at constant pressure is $\frac{7}{2}R$.

Write the expression to calculate the heat exhausted from the gas.

  $\left|Q^{\prime }\right|=\left|n{c}_p\left(T-T^{\prime }\right)\right|$

Here, $Q^{\prime }$ is the heat exhausted from the gas.

Substitute $\frac{7}{2}R$ for $c_p$ in the above equation to rewrite.

  $\left|Q^{\prime }\right|=\left|n\left(\frac{7}{2}R\right)\left(T-T^{\prime }\right)\right|$

Conclusion:

Substitute $2.00\text{mol}$ for n and $8.314\text{J}/\text{mol}\cdot \text{K}$ for R, $27.4\text{K}$ for T and $82.2\text{K}$ for $T^{\prime }$ in the above equation to calculate $Q^{\prime }$.

    $\begin{array}{c}\left|Q^{\prime }\right|=\left|\left(2.00\text{mol}\right)\left(\frac{7}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(27.4\text{K}-82.2\text{K}\right)\right|\\ =3.19\times {10}^3\text{J}\end{array}$

Therefore, the energy exhausted from the gas is $3.19\times {10}^3\text{J}$.

To determine

The efficiency of the cycle.

Answer to Problem 73PQ

The efficiency of the cycle is $12.9\%$.

Explanation of Solution

Write the expression to calculate the efficiency of one cycle.

  $e=\left(\frac{\left|W\right|}{\left|Q^{\prime }\right|}\right)\times 100\%$

Here, e is the efficiency.

Conclusion:

Substitute $3.19\times {10}^3\text{J}$ for $Q^{\prime }$ and $411\text{J}$ for W in the above equation to calculate e.

  $\begin{array}{c}e=\left(\frac{\left|411\text{J}\right|}{\left|3.19\times {10}^3\text{J}\right|}\right)\times 100\%\\ =12.9\%\end{array}$

Therefore, the efficiency of the cycle is $12.9\%$.

To determine

The efficiency of the Carnot engine.

Answer to Problem 73PQ

The efficiency of the Carnot engine is $66.7\%$.

Explanation of Solution

Write the expression to calculate the efficiency of the Carnot engine.

  $e^{\prime }=\left(1-\frac{T}{T^{\prime }}\right)\times 100\%$

Here, $e^{\prime }$ is the efficiency of the Carnot engine.

Conclusion:

Substitute $27.4\text{K}$ for T and $82.2\text{K}$ for $T^{\prime }$ in the above equation to calculate $e^{\prime }$.

  $\begin{array}{c}{e}^{\prime }=\left(1-\frac{27.4\text{K}}{82.2\text{K}}\right)\times 100\%\\ =66.7\%\end{array}$

Therefore, the efficiency of the Carnot engine is $66.7\%$.