Chapter 19, Problem 31P

(a)

To determine

The P - Vdiagram of the cycle, work done by the gas, heat absorbed by the gas and change in internal energy of the gas.

Answer to Problem 31P

The work done by the gas in each step are $0\text{J}$ , $4970\text{J}$ , $0\text{J}$ , and $-2484\text{J}$ . The heat absorbed by the gas in each step are $2486\text{J}$ , $12400\text{J}$ , $-4960\text{J}$ , and $-2490\text{J}$ and the change in internal energy in each step are $2486\text{J}$ , $7430\text{J}$ , $-4960\text{J}$ , and $-2490\text{J}$ .

Explanation of Solution

Given:

The initial pressure of the gas is $P_1=1.00\text{atm}$ .

The initial volume of the gas is $V_1=24.6\text{L}$ .

The pressure at state 2 and 3 is $P_{2,3}=2.00\text{atm}$ .

The volume at state 3 and 4 is $V_{3,4}=49.2\text{L}$ .

The pressure final pressure of the gas is $P_4=1.00\text{atm}$ .

Formula used:

The expression for the specific heat ratio is given as,

  $\gamma =\frac{c_p}{c_v}$

Here, $c_p$ is the specific heat at constant pressure and $c_v$ is the specific heat at constant volume.

The expression for heat absorbed by the gas in stage 1-2 is given as,

  $Q_{1-2}=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$

The expression for heat absorbed by the gas in stage 2-3 is given as,

  $Q_{2-3}=\frac{\gamma }{\gamma -1}\left(P_3{V}_3-P_2{V}_2\right)$

The expression for heat absorbed by the gas in stage 3-4 is given as,

  $Q_{3-4}=\frac{P_4{V}_4-P_3{V}_3}{\gamma -1}$

The expression for heat absorbed by the gas in stage 4-1 is given as,

  $Q_{4-1}=\frac{\gamma }{\gamma -1}\left(P_1{V}_1-P_4{V}_4\right)$

The expression for the work done in stage 2-3 is given as,

  $w_{2-3}=P_2\left(V_3-V_2\right)$

The expression for the work done in stage 4-1 is given as,

  $w_{4-1}=P_1\left(V_1-V_4\right)$

The expression for change in internal energy in stage 1-2 is given as,

  $d{u}_{1-2}=Q_{1-2}-w_{1-2}$

The expression for change in internal energy in stage 2-3 is given as,

  $d{u}_{2-3}=Q_{2-3}-w_{2-3}$

The expression for change in internal energy in stage 3-4 is given as,

  $d{u}_{3-4}=Q_{3-4}-w_{3-4}$

The expression for change in internal energy in stage 4-1 is given as,

  $d{u}_{4-1}=Q_{4-1}-w_{4-1}$

Calculation:

The P-V diagram for the cycle can be given as,

  

Figure 1

For monoatomic gas specific heat at constant volume and constant pressure can be given as,

  $\begin{array}{l}{c}_p=20.8\text{J}/\text{mol}\cdot \text{K}\\ c_v=12.5\text{J}/\text{mol}\cdot \text{K}\end{array}$

The specific heat ratio can be calculated as,

  $\begin{array}{c}\gamma =\frac{20.8\text{J}/\text{mol}\cdot \text{K}}{12.5\text{J}/\text{mol}\cdot \text{K}}\\ =1.67\end{array}$

The heat absorbed by the gas in stage 1-2 can be calculated as,

  $\begin{array}{c}{Q}_{1-2}=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}\\ =\frac{\left(2\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(24.6\text{L}\right)-\left(1\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)24.6}{1.67-1}\\ =\frac{49.7\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}-24.84\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}}{0.67}\\ =2486\text{J}\end{array}$

The expression for heat absorbed by the gas in stage 2-3 is given as,

  $\begin{array}{c}{Q}_{2-3}=\frac{\gamma }{\gamma -1}\left(P_3{V}_3-P_2{V}_2\right)\\ =\frac{1.67}{1.67-1}\left(\left(2\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(49.2\text{L}\right)-\left(2\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(24.6\text{L}\right)\right)\\ =2.5\left(\left(99.3\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\right)-\left(49.7\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\right)\right)\\ =12400\text{J}\end{array}$

The expression for heat absorbed by the gas in stage 3-4 is given as,

  $\begin{array}{c}{Q}_{3-4}=\frac{P_4{V}_4-P_3{V}_3}{\gamma -1}\\ =\frac{\left(1\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(49.2\text{L}\right)-\left(2\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(49.2\text{L}\right)}{1.67-1}\\ =\frac{49.7\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}-99.3\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}}{0.67}\\ =-4960\text{J}\end{array}$

The expression for heat absorbed by the gas in stage 4-1 is given as,

  $\begin{array}{c}{Q}_{4-1}=\frac{\gamma }{\gamma -1}\left(P_1{V}_1-P_4{V}_4\right)\\ =\frac{1.67}{1.67-1}\left(\left(1\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(24.6\text{L}\right)-\left(1\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\right)\left(49.2\text{L}\right)\right)\\ =2.5\left(\left(24.8\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\right)-\left(49.7\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\right)\right)\\ =-2490\text{J}\end{array}$

The work done in stage 1-2 is zero because it is constant volume process.

The work done in stage 2-3 can be calculated as,

  $\begin{array}{c}{w}_{2-3}=P_2\left(V_3-V_2\right)\\ =2\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\left(49.2\text{L}-24.6\text{L}\right)\\ =49.7\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\\ =4970\text{J}\end{array}$

The work done in stage 3-4 is zero because it is constant volume process.

The work done in stage 4-1 can be calculated as,

  $\begin{array}{c}{w}_{4-1}=P_1\left(V_1-V_4\right)\\ =1\text{atm}\times \frac{1.01\text{bar}}{1\text{atm}}\left(24.6\text{L}-49.2\text{L}\right)\\ =-24.84\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}\\ =-2484\text{J}\end{array}$

The expression for change in internal energy in stage 1-2 is given as,

  $\begin{array}{c}d{u}_{1-2}=Q_{1-2}-w_{1-2}\\ =2486\text{J}-0\\ =2486\text{J}\end{array}$

The expression for change in internal energy in stage 2-3 is given as,

  $\begin{array}{c}d{u}_{2-3}=Q_{2-3}-w_{2-3}\\ =12400\text{J}-4970\text{J}\\ =7430\text{J}\end{array}$

The expression for change in internal energy in stage 3-4 is given as,

  $\begin{array}{c}d{u}_{3-4}=Q_{3-4}-w_{3-4}\\ =-4960\text{J}-0\\ =-4960\text{J}\end{array}$

The expression for change in internal energy in stage 4-1 is given as,

  $\begin{array}{c}d{u}_{4-1}=Q_{4-1}-w_{4-1}\\ =-2490\text{J}-0\\ =-2490\text{J}\end{array}$

Conclusion:

Therefore,the work done by the gas in each step are $0\text{J}$ , $4970\text{J}$ , $0\text{J}$ , and $-2484\text{J}$ . Heat absorbed by the gas in each step are $2486\text{J}$ , $12400\text{J}$ , $-4960\text{J}$ , and $-2490\text{J}$ .Change in internal energy in each step are $2486\text{J}$ , $7430\text{J}$ , $-4960\text{J}$ , and $-2490\text{J}$ .

(b)

To determine

The efficiency of the cycle.

Answer to Problem 31P

The efficiency of the cycle is $16.7\%$ .

Explanation of Solution

Formula used:

The expression for total heat supplied to the cycle is given as,

  $Q_s=Q_{12}+Q_{23}$

The expression for total work done in the cycle is given as,

  $w=w_{1-2}+w_{2-3}+w_{3-4}+w_{4-1}$

The expression for efficiency of the cycle is given as,

  $\eta =\frac{w}{Q_s}\times 100$

Here, $Q_s$ is the total heat supplied to the cycle.

Calculation:

The total heat supplied to the cycle can be calculated as,

  $\begin{array}{c}{Q}_s=Q_{12}+Q_{23}\\ =2486\text{J}+12400\text{J}\\ =14886\text{J}\end{array}$

The total work done in the cycle can be calculated as,

  $\begin{array}{c}w=w_{1-2}+w_{2-3}+w_{3-4}+w_{4-1}\\ =0+4970\text{J}+0-2484\text{J}\\ =2486\text{J}\end{array}$

The efficiency of the cycle can be calculated as,

  $\begin{array}{c}\eta =\frac{w}{Q_s}\times 100\\ =\frac{2486\text{J}}{14886\text{J}}\times 100\\ =16.7\%\end{array}$

Conclusion:

Therefore, the efficiency of the cycle is $16.7\%$ .