Chapter 22, Problem 22.81CP

A 1.00-mol sample of an ideal gas (γ = 1.40) is carried through the Carnot cycle described in Figure 22.11. At point A, the pressure is 25.0 atm and the temperature is 600 K. At point C, the pressure is 1.00 atm and the temperature is 400 K. (a) Determine the pressures and volumes at points A, B, C, and D. (b) Calculate the net work done per cycle.

To determine

The pressure and volume at points $\text{A}$, $\text{B}$, $\text{C}$ and $\text{D}$.

Answer to Problem 22.81CP

For the point $\text{A}$, the pressure and volume are $25.0\text{atm}$ and $1.97\times {10}^{-3}{\text{m}}^{\text{3}}$. For the point $\text{B}$, the pressure and volume are $4.137\text{atm}$ and $1.19\times {10}^{-2}{\text{m}}^{\text{3}}$. For the point $\text{C}$, the pressure and volume are $1.00\text{atm}$ and $3.28\times {10}^{-2}{\text{m}}^{\text{3}}$. For the point $\text{D}$, the pressure and volume are $6.044\text{atm}$ and $5.43\times {10}^{-3}{\text{m}}^{\text{3}}$.

Explanation of Solution

Given data: The number of mole is $1$, the ratio of specific heat is $1.4$, the pressure and temperature at point $\text{A}$ are $25.0\text{atm}$ and $600\text{K}$ simultaneously and the pressure and temperature at point $\text{C}$ are $1.00\text{atm}$ and $400\text{K}$ simultaneously.

Consider the figure of Carnot cycle,

Figure (1)

From the figure, it is clear that the process from $\text{A}$ to $\text{B}$ and $\text{C}$ to $\text{D}$ is isothermal and the process from $\text{B}$ to $\text{C}$ and $\text{D}$ to $\text{A}$ is adiabatic.

Apply the ideal gas equation for volume at $\text{A}$,

$V_{\text{A}}=\frac{nR{T}_{\text{A}}}{P_{\text{A}}}$ (1)

Here,

$P_{\text{A}}$ is the pressure at point $\text{A}$.

$T_{\text{A}}$ is the temperature at point $\text{A}$.

$n$ is the number of mole of gas.

$R$ is the ideal gas constant.

The value of ideal gas constant is $8.314\text{J}/\text{mol}\text{.K}$.

Substitute $25.0\text{atm}$ for $P_{\text{A}}$, $600\text{K}$ for $T_{\text{A}}$, $8.314\text{J}/\text{mol}\text{.K}$ for $R$ and $1$ for $n$ in equation (1).

$\begin{array}{c}{V}_{\text{A}}=\frac{1\times 8.314\text{J}/\text{mol}\text{.K}\times 600\text{K}}{25.0\text{atm}\left(\frac{101325\text{Pa}}{1\text{atm}}\right)}\\ =1.97\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume at point $\text{A}$ is $1.97\times {10}^{-3}{\text{m}}^{\text{3}}$.

Apply the ideal gas equation for volume at $\text{C}$,

$V_{\text{C}}=\frac{nR{T}_{\text{C}}}{P_{\text{C}}}$ (2)

Here,

$P_{\text{C}}$ is the pressure at point $\text{C}$.

$T_{\text{C}}$ is the temperature at point $\text{C}$.

Substitute $1.00\text{atm}$ for $P_{\text{C}}$, $400\text{K}$ for $T_{\text{C}}$, $8.314\text{J}/\text{mol}\text{.K}$ for $R$ and $1$ for $n$ in equation (2).

$\begin{array}{c}{V}_{\text{C}}=\frac{1\times 8.314\text{J}/\text{mol}\text{.K}\times 400\text{K}}{1.00\text{atm}\left(\frac{101325\text{Pa}}{1\text{atm}}\right)}\\ =3.28\times {10}^{-2}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume at point $\text{C}$ is $3.28\times {10}^{-2}{\text{m}}^{\text{3}}$.

For the adiabatic process:

The condition is,

$\begin{array}{c}{T}_{\text{B}}{V}_{\text{B}}^{\gamma -1}=T_{\text{C}}{V}_{\text{C}}^{\gamma -1}\\ \frac{V_{\text{B}}^{\gamma -1}}{V_{\text{C}}^{\gamma -1}}=\frac{T_{\text{C}}}{T_{\text{B}}}\\ V_{\text{B}}=V_{\text{C}}{\left(\frac{T_{\text{C}}}{T_{\text{B}}}\right)}^{\frac{1}{\gamma -1}}\end{array}$ . (3)

Here,

$T_{\text{B}}$ is the temperature at point $\text{B}$.

$V_{\text{B}}$ is the volume of gas at point $\text{B}$.

$T_{\text{C}}$ is the temperature at point $\text{C}$.

$V_{\text{C}}$ is the volume of gas at point $\text{C}$.

$\gamma$ is the ratio of specific heat.

Substitute $3.28\times {10}^{-2}{\text{m}}^{\text{3}}$ for $V_{\text{C}}$, $400\text{K}$ for $T_{\text{C}}$, $600\text{K}$ for $T_{\text{B}}$ and $1.4$ for $\gamma$ in equation (3).

$\begin{array}{c}{V}_{\text{B}}=3.28\times {10}^{-2}{\text{m}}^{\text{3}}\times {\left(\frac{400\text{K}}{600\text{K}}\right)}^{\frac{1}{1.4-1}}\\ =3.28\times {10}^{-2}{\text{m}}^{\text{3}}\times 0.36289\\ =1.19\times {10}^{-2}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of gas at point $\text{B}$ is $1.19\times {10}^{-2}{\text{m}}^{\text{3}}$.

The formula for the pressure at point $\text{B}$ is,

$P_{\text{B}}=\frac{nR{T}_{\text{B}}}{V_{\text{B}}}$

Substitute $1.19\times {10}^{-2}{\text{m}}^{\text{3}}$ for $V_{\text{B}}$, $600\text{K}$ for $T_{\text{B}}$, $8.314\text{J}/\text{mol}\text{.K}$ for $R$ and $1$ for $n$ in above equation.

$\begin{array}{c}{P}_{\text{B}}=\frac{1\times 8.314\text{J}/\text{mol}\text{.K}\times 600\text{K}}{1.19\times {10}^{-2}{\text{m}}^{\text{3}}}\\ =419193.28\text{pa}\times \left(\frac{1\text{atm}}{101325\text{Pa}}\right)\\ =4.137\text{atm}\end{array}$

Thus, the pressure at point $\text{B}$ is $4.137\text{atm}$.

For the adiabatic process:

The condition is,

$\begin{array}{c}{T}_{\text{D}}{V}_{\text{D}}^{\gamma -1}=T_{\text{A}}{V}_{\text{A}}^{\gamma -1}\\ \frac{V_{\text{D}}^{\gamma -1}}{V_{\text{A}}^{\gamma -1}}=\frac{T_{\text{A}}}{T_{\text{D}}}\\ V_{\text{D}}=V_{\text{A}}{\left(\frac{T_{\text{A}}}{T_{\text{D}}}\right)}^{\frac{1}{\gamma -1}}\end{array}$

Here,

$T_{\text{D}}$ is the temperature at point $\text{D}$.

$V_{\text{D}}$ is the volume of gas at point $\text{D}$.

Substitute $1.97\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_{\text{A}}$, $400\text{K}$ for $T_{\text{D}}$, $600\text{K}$ for $T_{\text{A}}$ and $1.4$ for $\gamma$ in above equation.

$\begin{array}{c}{V}_{\text{D}}=1.97\times {10}^{-3}{\text{m}}^{\text{3}}\times {\left(\frac{600\text{K}}{400\text{K}}\right)}^{\frac{1}{1.4-1}}\\ =1.97\times {10}^{-3}{\text{m}}^{\text{3}}\times 2.76\\ =5.43\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of gas at point $\text{D}$ is $5.43\times {10}^{-3}{\text{m}}^{\text{3}}$.

The formula for the pressure at point $\text{D}$ is,

$P_{\text{D}}=\frac{nR{T}_{\text{D}}}{V_{\text{D}}}$

Substitute $5.43\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_{\text{D}}$, $400\text{K}$ for $T_{\text{D}}$, $8.314\text{J}/\text{mol}\text{.K}$ for $R$ and $1$ for $n$ in above equation.

$\begin{array}{c}{P}_{\text{D}}=\frac{1\times 8.314\text{J}/\text{mol}\text{.K}\times 400\text{K}}{5.43\times {10}^{-3}{\text{m}}^{\text{3}}}\\ =612449.35\text{pa}\times \left(\frac{1\text{atm}}{101325\text{Pa}}\right)\\ =6.044\text{atm}\end{array}$

Conclusion:

Therefore, For the point $\text{A}$, the pressure and volume are $25.0\text{atm}$ and $1.97\times {10}^{-3}{\text{m}}^{\text{3}}$. For the point $\text{B}$, the pressure and volume are $4.137\text{atm}$ and $1.19\times {10}^{-2}{\text{m}}^{\text{3}}$. For the point $\text{C}$, the pressure and volume are $1.00\text{atm}$ and $3.28\times {10}^{-2}{\text{m}}^{\text{3}}$. For the point $\text{D}$, the pressure and volume are $6.044\text{atm}$ and $5.43\times {10}^{-3}{\text{m}}^{\text{3}}$.

To determine

The net work done per cycle.

Answer to Problem 22.81CP

The net work done per cycle is $2.99\times {10}^{-3}\text{J}$.

Explanation of Solution

The formula for the net work done per cycle is,

$w=nR\left(T_{\text{A}}-T_{\text{C}}\right)\ln \left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)$

Substitute $1$ for $n$, $8.314\text{J}/\text{mol}\text{.K}$ for $R$, $600\text{K}$ for $T_{\text{A}}$, $400\text{K}$ for $T_{\text{C}}$, $1.19\times {10}^{-2}{\text{m}}^{\text{3}}$ for $V_{\text{B}}$ and $1.97\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V_{\text{A}}$ in above equation.

$\begin{array}{c}w=1\times 8.314\text{J}/\text{mol}\text{.K}\times \left(600\text{K}-400\text{K}\right)\ln \left(\frac{1.19\times {10}^{-2}{\text{m}}^{\text{3}}}{1.97\times {10}^{-3}{\text{m}}^{\text{3}}}\right)\\ =2.99\times {10}^{-3}\text{J}\end{array}$

Conclusion:

Therefore, the net work done per cycle is $2.99\times {10}^{-3}\text{J}$.