Chapter 19, Problem 34P

(a)

To determine

The temperature of each numbered state of the cycle.

Answer to Problem 34P

The temperature at state 1 is $301\text{K}$ , the temperature at state 2 is $601\text{K}$ and the temperature at state 1 is $601\text{K}$ .

Explanation of Solution

Given:

The number of mol of an ideal monatomic gas is $n=1\text{mol}$ .

The initial volume of gas is $V_1=25.0\text{L}$ .

Formula used:

The expression for ideal gas law is given as,

  $PV=nRT$

Here, $R$ is the gas constant and $T$ is the temperature.

Calculation:

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

The temperature $T_1$ is calculated as,

  $\begin{array}{c}{T}_1=\frac{P_1{V}_1}{nR}\\ =\frac{\left(100\text{kPa}\right)\left(25.0\text{L}\right)}{1.00\text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}}\\ =301\text{K}\end{array}$

The temperature $T_2$ is calculated as,

  $\begin{array}{c}{T}_2=\frac{P_2{V}_1}{nR}\\ =\frac{\left(200\text{kPa}\right)\left(25.0\text{L}\right)}{\left(1.00\text{mol}\right)\times 8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}}\\ =601\text{K}\end{array}$

The temperature $T_3$ is calculated as,

  $\begin{array}{c}{T}_3=T_2\\ =601\text{K}\end{array}$

Conclusion:

Therefore, the temperature at state 1 is $301\text{K}$ , the temperature at state 2 is $601\text{K}$ and the temperature at state 1 is $601\text{K}$ .

(b)

To determine

The heat transfer for each part of the cycle.

Answer to Problem 34P

The heat transfer for process $1\to 2$ is $3.74\text{kJ}$ , the heat transfer for process $2\to 3$ is $3.46\text{kJ}$ and the heat transfer for process $3\to 1$ is $-6.24\text{kJ}$ .

Explanation of Solution

Formula used:

The expression for heat transfer for constant volume process $1\to 2$ is given as,

  $Q_{12}=\frac{3}{2}R\Delta T_{12}$

Here, R is the gas constant.

The expression for heat transfer during isothermal process $2\to 3$ is given as,

  $Q_{23}=nR{T}_2\ln \left(\frac{V_3}{V_2}\right)$

The expression for heat transfer during isobaric compression process $3\to 1$ is given as,

  $Q_{31}=\frac{5}{2}R\Delta T_{31}$

Calculation:

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

The heat transfer for constant volume process $1\to 2$ is calculated as,

  $\begin{array}{c}{Q}_{12}=\frac{3}{2}R\Delta T_{12}\\ =\frac{3}{2}\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(601\text{K}-301\text{K}\right)\\ =3.74\text{kJ}\end{array}$

The heat transfer for isothermal process $2\to 3$ is calculated as,

  $\begin{array}{c}{Q}_{23}=nR{T}_2\ln \left(\frac{V_3}{V_2}\right)\\ =\left(1.00\text{mol}\right)\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(601\text{K}\right)\ln \left(\frac{2\times 25}{25}\right)\\ =3.46\text{kJ}\end{array}$

The heat transfer for isobaric compression process $3\to 1$ is calculated as,

  $\begin{array}{c}{Q}_{31}=\frac{5}{2}R\Delta T_{31}\\ =\frac{5}{2}\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(301\text{K}-601\text{K}\right)\\ =-6.24\text{kJ}\end{array}$

Conclusion:

Therefore, the heat transfer for process $1\to 2$ is $3.74\text{kJ}$ , the heat transfer for process $2\to 3$ is $3.46\text{kJ}$ and the heat transfer for process $3\to 1$ is $-6.24\text{kJ}$ .

(c)

To determine

The efficiency of the cycle.

Answer to Problem 34P

The efficiency of the cycle is $13\%$ .

Explanation of Solution

Formula used:

The expression for the efficiency of the cycle is given as,

  $\epsilon =\frac{W}{Q_{in}}$

The expression for heat addition is given as,

  $Q_{in}=Q_{12}+Q_{23}$

The expression for work done from first law of thermodynamics is given as,

  $W={\displaystyle \sum Q}=Q_{12}+Q_{23}+Q_{31}$

Calculation:

The heat addition is calculated as,

  $\begin{array}{c}{Q}_{in}=Q_{12}+Q_{23}\\ =3.74\text{kJ}+3.46\text{kJ}\\ =7.2\text{kJ}\end{array}$

The work done is calculated as,

  $\begin{array}{c}W={\displaystyle \sum Q}=Q_{12}+Q_{23}+Q_{31}\\ =3.74\text{kJ}+3.46\text{kJ}-6.24\text{kJ}\\ =0.96\text{kJ}\end{array}$

The efficiency of the cycle is calculated as,

  $\begin{array}{c}\epsilon =\frac{W}{Q_{in}}\\ =\frac{0.96\text{kJ}}{7.2\text{kJ}}\\ =13\%\end{array}$

Conclusion:

Therefore, the efficiency of the cycle is $13\%$ .