Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is $2.00\text{atm}$ .

The initial volume is $2.00\text{L}$

The final volume is $4.00\text{L}$

Calculation:

Consider the given data, the pressure versus volume graph for the process is shown below.

  

Figure 1

Conclusion:

Therefore, the P-V diagram is shown in figure 1.

(b)

To determine

The temperature at end of each part of cycle.

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are $24.4\text{K}$ , $24.4\text{K}$ and $48.7\text{K}$ respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

  $T_1=\frac{P_1{V}_1}{nR}$

The expression for temperature for portion 2 is given by,

  $T_2=T_1$

The expression for temperature for portion 3 is given by,

  $T_3=\frac{P_3{V}_3}{nR}$

Calculation:

The temperature for portion1 is calculated as,

  $\begin{array}{c}{T}_1=\frac{P_1{V}_1}{nR}\\ =\frac{\left(2.00\text{atm}\right)\left(2\text{L}\right)}{\left(\left(2.00\text{mole}\right)\left(0.08206\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\right)\right)}\\ =24.4\text{K}\end{array}$

The temperature for portion2 is calculated as,

  $\begin{array}{c}{T}_2=T_1\\ =24.4\text{K}\end{array}$

The temperature for portion3 is calculated as,

  $\begin{array}{c}{T}_3=\frac{P_3{V}_3}{nR}\\ =\frac{\left(2.00\text{atm}\right)\left(4\text{L}\right)}{\left(\left(2.00\text{mole}\right)\left(0.08206\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\right)\right)}\\ =48.7\text{K}\end{array}$

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are $24.4\text{K}$ , $24.4\text{K}$ and $48.7\text{K}$ respectively.

(c)

To determine

The head absorbed and work done during each cycle.

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are $281\text{J}$ , $606\text{J}$ and $1010.15\text{J}$ respectively. The work done for process 1, 2 and 3 are $281\text{J}$ , $0$ and $405\text{J}$ respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

  $Q_{\text{in1}}=nRT\ln \frac{V_2}{V_1}$

The expression for heat absorbed for portion 2 is given by,

  $Q_{\text{in2}}=\frac{3}{4}nR\left(T_3-T_2\right)$

The expression for heat absorbed for portion 3 is given by,

  $Q_{\text{in3}}=\frac{5}{2}nR\left(T_1-T_3\right)$

The expression for work for portion 1 is given by,

  $W_1=Q_{\text{in1}}$

The expression for work done for portion 3 is given by,

  $W_3=-P_{1,3}\left(V_1-V_3\right)$

Calculation:

The heat absorbed for portion1 is calculated as,

  $\begin{array}{c}{Q}_{\text{in1}}=nRT\ln \frac{V_2}{V_1}\\ =\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(24.4\text{K}\right)\ln \left(\frac{\text{4}\text{L}}{2\text{L}}\right)\\ =281\text{J}\end{array}$

The work done for portion1 is calculated as,

  $\begin{array}{c}{W}_1=Q_{\text{in1}}\\ =281\text{J}\end{array}$

The heat absorbed for portion2 is calculated as,

  $\begin{array}{c}{Q}_{\text{in2}}=\frac{3}{4}nR\left(T_3-T_2\right)\\ =\frac{3}{2}\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(48.7\text{K}-24.4\text{K}\right)\\ =606\text{J}\end{array}$

The heat absorbed for portion3 is calculated as,

  $\begin{array}{c}{Q}_{\text{in3}}=\frac{5}{2}nR\left(T_1-T_3\right)\\ =\frac{5}{2}\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(48.7\text{K}-24.4\text{K}\right)\\ =1010.15\text{J}\end{array}$

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

  $\begin{array}{c}{W}_3=-P_{1,3}\left(V_1-V_3\right)\\ =-\left(2.00\text{atm}\right)\left(\text{2}\text{L}-4\text{L}\right)\\ =-\left(\left(-4\text{atm}\cdot \text{L}\right)\left(\frac{101.3\text{J}}{1\text{atm}\cdot \text{L}}\right)\right)\\ =405\text{J}\end{array}$

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are $281\text{J}$ , $606\text{J}$ and $1010.15\text{J}$ respectively. The work done for process 1, 2 and 3 are $281\text{J}$ , $0$ and $405\text{J}$ respectively.