Chapter 19, Problem 47P

(a)

To determine

The temperature after the adiabatic expansion.

Answer to Problem 47P

The temperature after the adiabatic expansion is $373\text{K}$ .

Explanation of Solution

Given:

The initial pressure of gas is $P_1=1\text{atm}$ .

The initial temperature is $T_1=0°\text{C}$ .

The final temperature is $T_2=150°\text{C}$ .

Formula used:

The temperature after the adiabatic expansion is given as,

  $T_3=T_1\times \frac{V_3}{V_1}$

Here, $V_3$ is the volume after adiabatic expansion and $V_1$ is the initial volume.

The volume after adiabatic expansion is given as,

  $V_3=V_2\times {\left(\frac{P_2}{P_3}\right)}^{\frac{1}{\gamma }}$

Here, $V_2$ is the volume after adiabatic expansion, $P_3$ is the pressure at adiabatic expansion, $P_2$ is the final pressure and $\gamma$ is the adiabatic index. and its value is 1.41.

The final pressure is given as,

  $P_2=P_1\times \frac{T_2}{T_1}$

Calculation:

The volume occupied by the gas at $1\text{atm}$ is $22.4\text{L}$ . Then the initial volume and final volume is calculated as,

  $\begin{array}{l}{V}_1=22.4\text{L}\\ V_1=V_2\\ V_2=22.4\text{L}\end{array}$

For the fixed amount of gas the pressure at adiabatic expansion is can be calculated as,

  $\begin{array}{l}{P}_1=P_3\\ P_1=1\text{atm}\\ P_3=1\text{atm}\end{array}$

The final pressure is calculated as,

  $\begin{array}{c}{P}_2=P_1\times \frac{T_2}{T_1}\\ =1\text{atm}\times \frac{150°\text{C}+273\text{K}}{0°\text{C}+273\text{K}}\\ =1\text{atm}\times \frac{423\text{K}}{273\text{K}}\\ =1.55\text{atm}\end{array}$

The volume after adiabatic expansion is calculated as,

  $\begin{array}{c}{V}_3=V_2\times {\left(\frac{P_2}{P_3}\right)}^{\frac{1}{\gamma }}\\ =22.4\text{L}\times {\left(\frac{1.55\text{atm}}{1\text{atm}}\right)}^{\frac{1}{1.41}}\\ =30.6\text{L}\end{array}$

The temperature after the adiabatic expansion is calculated as,

  $\begin{array}{c}{T}_3=T_1\times \frac{V_3}{V_1}\\ =0°\text{C}+273\text{K}\times \frac{30.6\text{L}}{22.4\text{L}}\\ =273\text{K}\times \frac{30.6\text{L}}{22.4\text{L}}\\ =373\text{K}\end{array}$

Conclusion:

Therefore, the temperature after the adiabatic expansion is $373\text{K}$ .

(b)

To determine

The heat absorbed or released by the system during each step.

Answer to Problem 47P

The heat energy absorbed during the process 1-2 is $3.12\text{kJ}$ , the heat energy during the process 2-3 is zero and the heat energy released during the process 3-1 is $-2.91\text{kJ}$ .

Explanation of Solution

Formula used:

The heat energy during the constant volume process 1-2 is given as,

  $Q_{1-2}=C_v\left(T_2-T_1\right)$

Here, $C_v$ is the specific heat at constant volume and its value for the diatomic gas is given as,

  $C_v=\frac{5}{2}R$

Here, $R$ is gas constant and its value is $8.314\text{J}/\text{mol}\cdot \text{K}$ .

The heat energy during the constant pressure process 3-1 is given as,

  $Q_{1-2}=C_p\left(T_1-T_3\right)$

Here, $C_p$ is the specific heat at constant pressure and its value for the diatomic gas is given as,

  $C_p=\frac{7}{2}R$

Calculation:

The heat energy during the constant volume process 1-2 is calculated as,

  $\begin{array}{c}{Q}_{1-2}=C_v\left(T_2-T_1\right)\\ =\frac{5}{2}\times R\times \left(T_2-T_1\right)\\ =\frac{5}{2}\times 8.314\text{J}/\text{mol}\cdot \text{K}\times \left[\left(150°\text{C}+273\text{K}\right)-\left(0°\text{C}+273\right)\text{K}\right]\\ =3.12\text{kJ}\end{array}$

The process 2-3 is an adiabatic process. The heat energy for the process 2-3 is calculated as,

  $Q_{2-3}=0$

The heat energy during the constant pressure process 3-1 is given as,

  $\begin{array}{c}{Q}_{3-1}=C_p\left(T_1-T_3\right)\\ =\frac{7}{2}\times R\times \left(T_2-T_1\right)\\ =\frac{7}{2}\times 8.314\text{J}/\text{mol}\cdot \text{K}\times \left[273\text{K}-373\text{K}\right]\\ =-2.91\text{kJ}\end{array}$

The negative sign shows that heat is released during the process.

Conclusion:

Therefore, the heat energy absorbed during the process 1-2 is $3.12\text{kJ}$ , the heat energy during the process 2-3 is zero and the heat energy released during the process 3-1 is $-2.91\text{kJ}$ .

(c)

To determine

The efficiency of the cycle.

Answer to Problem 47P

The efficiency of the cycle is $6.7\%$ .

Explanation of Solution

Formula used:

The efficiency of the cycle is given as,

  $\eta =\frac{W}{Q_{1-2}}$

Here, $W$ is the total work obtained during the cycle.

The total work during the cycle is given as,

  $W=Q_{1-2}+Q_{2-3}+Q_{3-1}$

Calculation:

The total work during the cycle is calculated as,

  $\begin{array}{c}W=Q_{1-2}+Q_{2-3}+Q_{3-1}\\ =3.12\text{kJ}+0-2.91\text{kJ}\\ =0.21\text{kJ}\end{array}$

The efficiency of the cycle is calculated as,

  $\begin{array}{c}\eta =\frac{W}{Q_{1-2}}\\ =\frac{0.21\text{kJ}}{3.12\text{kJ}}\\ =0.067\end{array}$

Further solving the above equation as,

  $\begin{array}{c}\eta =0.067\times 100\\ =6.7\%\end{array}$

Conclusion:

Therefore, the efficiency of the cycle is $6.7\%$ .

(d)

To determine

The efficiency of a Carnot cycle operating between the extreme temperatures.

Answer to Problem 47P

The efficiency of a Carnot cycle operating between the extreme temperatures is $35.4\%$ .

Explanation of Solution

Formula used:

The efficiency of the Carnot cycle at extreme temperature ( $423\text{K}$ and $273\text{K}$ ) is given as,

  $\eta =1-\frac{T_1}{T_2}$

Calculation:

The efficiency of the Carnot cycle at extreme temperature ( $423\text{K}$ and $273\text{K}$ ) can be calculated as,

  $\begin{array}{l}\eta =1-\frac{T_1}{T_2}\\ \eta =1-\frac{273\text{K}}{423\text{K}}\\ \eta =0.354\end{array}$

Further solving the above equation as,

  $\begin{array}{l}\eta =0.354\times 100\\ \eta =35.4\%\end{array}$

Conclusion:

Therefore, the efficiency of a Carnot cycle operating between the extreme temperatures is $35.4\%$ .