Chapter 19, Problem 78P

(a)

To determine

The volume of gas at A.

Answer to Problem 78P

The volume at A is $19.69\text{L}$ .

Explanation of Solution

Given:

The pressure at A is $P_{\text{A}}=5.00\text{atm}$ .

The temperature at A is $T_{\text{A}}=600\text{K}$ .

Formula used:

The expression for volume at A is given by,

  $V_{\text{A}}=\frac{nR{T}_{\text{A}}}{P_{\text{A}}}$

Calculation:

The volume at A is calculated as,

  $\begin{array}{c}{V}_{\text{A}}=\frac{nR{T}_{\text{A}}}{P_{\text{A}}}\\ =\frac{\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)}{\left(\left(5.00\text{atm}\right)\left(\frac{101.325\text{kPa}}{1\text{atm}}\right)\right)}\\ =\left(\left(1.969\times {10}^{-2}{\text{m}}^3\right)\left(\frac{{10}^3\text{L}}{1{\text{m}}^3}\right)\right)\\ =19.69\text{L}\end{array}$

Conclusion:

Therefore, the volume at A is $19.69\text{L}$ .

(b)

To determine

The volume and temperature of gas at B.

Answer to Problem 78P

The volume and temperature at B are $39.38\text{L}$ and $1200\text{K}$ respectively.

Explanation of Solution

Formula used:

The expression for volume at B is given by,

  $V_{\text{B}}=2V_{\text{A}}$

The expression for temperature at B is given by,

  $T_{\text{B}}=T_{\text{A}}\frac{V_{\text{B}}}{V_{\text{A}}}$

Calculation:

The volume at B is calculated as,

  $\begin{array}{c}{V}_{\text{B}}=2V_{\text{A}}\\ =2\left(19.69\text{L}\right)\\ =39.38\text{L}\end{array}$

The temperature at B is calculated as,

  $\begin{array}{c}{T}_{\text{B}}=T_{\text{A}}\frac{V_{\text{B}}}{V_{\text{A}}}\\ =\left(600K\right)\left(\frac{39.38\text{L}}{19.69\text{L}}\right)\\ =1200\text{K}\end{array}$

Conclusion:

Therefore, the volume and temperature at B are $39.38\text{L}$ and $1200\text{K}$ respectively.

(c)

To determine

The temperature of gas at C.

Answer to Problem 78P

The temperature at C is $600\text{K}$ .

Explanation of Solution

Formula used:

The expression for temperature at C as process is isothermal given by,

  $T_{\text{C}}=T_{\text{A}}$

Calculation:

The temperature at C is calculated as,

  $\begin{array}{c}{T}_{\text{C}}=T_{\text{A}}\\ =600\text{K}\end{array}$

Conclusion:

Therefore, the temperature at C is $600\text{K}$ .

(d)

To determine

The volume of gas at C.

Answer to Problem 78P

The volume at C is $222.77\text{L}$ .

Explanation of Solution

Formula used:

The expression for volume at C as process is given by,

  $V_{\text{C}}=V_{\text{B}}{\left(\frac{T_{\text{B}}}{T_{\text{C}}}\right)}^{\frac{1}{\gamma -1}}$

Calculation:

The volume at C is calculated as,

  $\begin{array}{c}{V}_{\text{C}}=V_{\text{B}}{\left(\frac{T_{\text{B}}}{T_{\text{C}}}\right)}^{\frac{1}{\gamma -1}}\\ =\left(39.38\text{L}\right){\left(\frac{1200\text{K}}{600\text{K}}\right)}^{\frac{1}{1.4-1}}\\ =222.77\text{L}\end{array}$

Conclusion:

Therefore, the volume at C is $222.77\text{L}$ .

(e)

To determine

The work done by gas in each segment of cycle.

Answer to Problem 78P

The work done in process AB, BC and CA are $9.98\text{kJ}$ , $24.94\text{kJ}$ and $-25.85\text{kJ}$ respectively.

Explanation of Solution

Formula used:

The work done in process AB is given by,

  $W_{\text{AB}}=P_{\text{A}}{V}_{\text{A}}$

The expression for work done in process BC is given by,

  $W_{\text{BC}}=-\frac{5}{2}nR\Delta T_{\text{BC}}$

The expression for work done in process CA is given by,

  $W_{\text{CA}}=nR{T}_{\text{C}}\ln \left(\frac{V_{\text{A}}}{V_{\text{C}}}\right)$

Calculation:

The work done in process AB is calculated as,

  $\begin{array}{c}{W}_{\text{AB}}=P_{\text{A}}{V}_{\text{A}}\\ =\left(5.00\text{atm}\right)\left(19.69\text{L}\right)\\ =\left(\left(98.45\text{atm}\cdot \text{L}\right)\left(\frac{0.101325\text{kJ}}{1\text{atm}\cdot \text{L}}\right)\right)\\ =9.98\text{kJ}\end{array}$

The work done in process BC is calculated as,

  $\begin{array}{c}{W}_{\text{BC}}=-\frac{5}{2}nR\Delta T_{\text{BC}}\\ =-\frac{5}{2}\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}-1200\text{K}\right)\\ =\left(\left(2.494\times {10}^4\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\right)\\ =24.94\text{kJ}\end{array}$

The work done in process CA is calculated as,

  $\begin{array}{c}{W}_{\text{CA}}=nR{T}_{\text{C}}\ln \left(\frac{V_{\text{A}}}{V_{\text{C}}}\right)\\ =\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)\ln \left(\frac{16.69\text{L}}{222.77\text{L}}\right)\\ =\left(\left(-25.85\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\right)\\ =-25.85\text{kJ}\end{array}$

Conclusion:

Therefore, the work done in process AB, BC and CA are $9.98\text{kJ}$ , $24.94\text{kJ}$ and $-25.85\text{kJ}$ respectively.

(f)

To determine

The heat absorbed in each segment of cycle.

Answer to Problem 78P

The heat absorbed in process AB, BC and CA are $34.92\text{kJ}$ , $0$ and $-25.85\text{kJ}$ respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed in process AB is given by,

  $Q_{\text{AB}}=n{c}_{\text{P}}\Delta T_{\text{AB}}$

The expression for heat absorbed in process CA is given by,

  $Q_{\text{CA}}=W_{\text{CA}}$

Calculation:

The heat absorbed in process AB is calculated as,

  $\begin{array}{c}{Q}_{\text{AB}}=n{c}_{\text{P}}\Delta T_{\text{AB}}\\ =\frac{7}{2}\left(2.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(1200\text{K}-600\text{K}\right)\\ =\left(\left(34.92\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\right)\\ =34.92\text{kJ}\end{array}$

The heat absorbed in process BC is 0 because the process is adiabatic.

The heat absorbed in process CA is calculated as,

  $\begin{array}{c}{Q}_{\text{CA}}=W_{\text{CA}}\\ =-25.85\text{kJ}\end{array}$

Conclusion:

Therefore, the heat absorbed in process AB, BC and CA are $34.92\text{kJ}$ , $0$ and $-25.85\text{kJ}$ respectively.