Chapter 3, Problem 3.65E

Interpretation Introduction

Interpretation:

The expressions to calculate the work and heat for the four steps of a Carnot cycle are to be stated. The $w$ and $q$ for individual steps, total work, and heat of the cycle are to be calculated. The value of $\Delta S$ for the cycle undergoing reversible conditions is zero is to be shown.

Concept introduction:

The Carnot cycle represents the relationship between the efficiency of a steam engine and the temperatures. It was given by Carnot who stated that every engine gets the heat from a high-temperature reservoir. Some of the heat is converted to the work during the process. The Carnot cycle has four major steps:

Step 1: Reversible isothermal expansion $\left(q_1,w_1\right)$.

Step 2: Reversible adiabatic expansion $\left(q_2=0,w_2\right)$.

Step 3: Reversible isothermal compression $\left(q_3,w_3\right)$.

Step 4: Reversible adiabatic compression $\left(q_4=0,w_4\right)$.

Answer to Problem 3.65E

The expressions to calculate the work and heat for the four steps of a Carnot cycle are given below.

$\begin{array}{rll}{q}_1& =& nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ q_2& =& 0\\ q_3& =& nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ q_4& =& 0\end{array}$

$\begin{array}{rll}{w}_1& =& -nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ w_3& =& -nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ w_2& =& p\left(V_{\text{b}}-V_{\text{c}}\right)\\ w_4& =& p\left(V_{\text{d}}-V_{\text{a}}\right)\end{array}$

The $w$ and $q$ for each step in the cycle are $q_1=-1.71\times {10}^3\text{J}$, $q_3=-7.12\times {10}^2\text{J}$, $q_2=0$, $q_4=0$, $w_1=1.71\times {10}^3\text{J}$, $w_3=7.12\times {10}^2\text{J}$, $w_2=-1\text{J}$ and $w_4=-3\text{J}$.

The total work and heat of the cycle is $2.418\times {10}^3\text{J}$ and $-2.42\times {10}^3\text{J}$ respectively.

The value of $\Delta S=0$, has been shown.

Explanation of Solution

For the Carnot cycle, the first and the third step are reversible isothermal processes.

On the other hand, the second and the fourth step are reversible adiabatic processes. So, the expressions to calculate the heat for four steps can be written as shown below.

$\begin{array}{rll}{q}_1& =& nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ q_2& =& 0\\ q_3& =& nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ q_4& =& 0\end{array}$

The variables $q_1$ and $q_3$ are the heat under the reversible isothermal conditions.

From the first law of thermodynamics, work done, $w=-q$ for the isothermal process. Therefore, the expressions to calculate the work for first and the third steps can be written as shown below.

$\begin{array}{rll}{w}_1& =& -nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ w_3& =& -nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\end{array}$

The work done for reversible adiabatic processes is given by the formula given below.

$w=p\Delta V$

Therefore, the expressions to calculate the work for second and the fourth steps can be written as shown below.

$\begin{array}{rll}{w}_2& =& p\Delta V\\ w_4& =& p\Delta V\end{array}$

Assume $p=1\text{Pa}$, volume, $V$ from $\text{1}{\text{m}}^{\text{3}}$ to $\text{4}{\text{m}}^{\text{3}}$, temperature, $T=298\text{K}$, and the number of moles, $n=\text{1}\text{mol}$ to calculate the work and heat as shown below.

$\begin{array}{rll}{q}_1& =& nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ & =& 1\text{mol}\times \text{8}\text{.314}\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \frac{\text{1}{\text{m}}^{\text{3}}}{2{\text{m}}^{\text{3}}}\\ & =& -1.71\times {10}^3\text{J}\end{array}$

$\begin{array}{rll}{q}_3& =& nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ & =& 1\text{mol}\times \text{8}\text{.314}\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \frac{3{\text{m}}^{\text{3}}}{4{\text{m}}^{\text{3}}}\\ & =& -7.12\times {10}^2\text{J}\end{array}$

Total heat can be calculated as shown below.

$q_{\text{cycle}}=q_1+q_2+q_3+q_4$

Substitute the values in the above equation as shown below.

$\begin{array}{rll}{q}_{\text{cycle}}& =& q_1+q_2+q_3+q_4\\ & =& -1.71\times {10}^3\text{J}+0+-7.12\times {10}^2\text{J}+0\\ & =& -2.42\times {10}^3\text{J}\end{array}$

Work done for the first and third step can be calculated as shown below.

$\begin{array}{rll}{w}_1& =& -q_1\\ w_3& =& -q_3\end{array}$

Substitute the values in the above equation as shown below.

$\begin{array}{rll}{w}_1& =& -q_1\\ & =& -\left(-1.71\times {10}^3\text{J}\right)\\ & =& 1.71\times {10}^3\text{J}\end{array}$

$\begin{array}{rll}{w}_3& =& -q_3\\ & =& -\left(-7.12\times {10}^2\text{J}\right)\\ & =& 7.12\times {10}^2\text{J}\end{array}$

Work done for the second and fourth step can be calculated as shown below.

$\begin{array}{rll}{w}_2& =& p\Delta V\\ w_4& =& p\Delta V\end{array}$

Substitute the values in the above equation as shown below.

$\begin{array}{rll}{w}_2& =& p\Delta V\\ & =& p\left(V_{\text{b}}-V_{\text{c}}\right)\\ & =& 1\text{Pa}\left(2{\text{m}}^{\text{3}}-3{\text{m}}^{\text{3}}\right)\times \frac{1\text{J}}{1\text{Pa}{\text{m}}^{\text{3}}}\\ & =& -1\text{J}\end{array}$

$\begin{array}{rll}{w}_4& =& p\Delta V\\ & =& p\left(V_{\text{d}}-V_{\text{a}}\right)\\ & =& 1\text{Pa}\left(4{\text{m}}^{\text{3}}-1{\text{m}}^{\text{3}}\right)\times \frac{1\text{J}}{1\text{Pa}{\text{m}}^{\text{3}}}\\ & =& -3\text{J}\end{array}$

The work of the cycle is calculated by using the formula given below.

$w_{\text{cycle}}=w_1+w_2+w_3+w_4$

Substitute the values in the above equation as shown below.

$\begin{array}{rll}{w}_{\text{cycle}}& =& w_1+w_2+w_3+w_4\\ & =& \left(1.71\times {10}^3\text{J}\right)\text{+}\left(-1\text{J}\right)+\left(7.12\times {10}^2\text{J}\right)+\left(-3\text{J}\right)\\ & =& 2.418\times {10}^3\text{J}\end{array}$

The entropy change is calculated by the formula given below.

$\Delta S={\displaystyle \int \frac{d{q}_{\text{rev}}}{T}}$

The steps first and the third are reversible isothermal processes; so, the temperature is constant, and at a constant temperature, the heat change is constant. The values of changes in the energy, that is, $d{q}_1$ and $d{q}_3$ is equal to zero. For adiabatic processes; that is, the second and fourth step, the heat change is zero; that is, $q_2$ and $q_4$ is zero. Therefore, the entropy for the reversible condition of the Carnot cycle is zero.

Conclusion

The expressions to calculate the work and heat for the four steps of a Carnot cycle are given below.

$\begin{array}{rll}{q}_1& =& nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ q_2& =& 0\\ q_3& =& nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ q_4& =& 0\end{array}$

$\begin{array}{rll}{w}_1& =& -nRT\ln \frac{V_{\text{f, a}}}{V_{\text{i, b}}}\\ w_3& =& -nRT\ln \frac{V_{\text{f, c}}}{V_{\text{i, d}}}\\ w_2& =& p\left(V_{\text{b}}-V_{\text{c}}\right)\\ w_4& =& p\left(V_{\text{d}}-V_{\text{a}}\right)\end{array}$

The $w$ and $q$ for each step in the cycle are $q_1=-1.71\times {10}^3\text{J}$, $q_3=-7.12\times {10}^2\text{J}$, $q_2=0$, $q_4=0$, $w_1=1.71\times {10}^3\text{J}$, $w_3=7.12\times {10}^2\text{J}$, $w_2=-1\text{J}$, and $w_4=-3\text{J}$.

The total work and heat of the cycle is $2.418\times {10}^3\text{J}$ and $-2.42\times {10}^3\text{J}$ respectively.

The value of $\Delta S=0$, has been shown.