Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Answer to Problem 75P

The total work done is $6.62\text{kJ}$ . The heat absorbed at point A, B, C and D are $13.2\text{kJ}$ , $-8.98\text{kJ}$ , $-6.58\text{kJ}$ and $8.98\text{kJ}$ respectively.

Explanation of Solution

Given:

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

The temperature is $360\text{K}$

Formula used:

The expression for volume at D is given by,

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

The expression for pressure at point C is given by,

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

The expression for temperature at point A and B is given by,

  $T_{\text{A}}=\frac{P_{\text{A}}{V}_{\text{A}}}{nR}$

The expression for heat absorbed at D is given by,

  $Q_{\text{D}}=\frac{3}{2}nR\Delta T_{\text{D}}$

The expression for heat absorbed at A is given by,

  $Q_{\text{A}}=nR{T}_{\text{A}}\ln \frac{V_{\text{B}}}{V_{\text{A}}}$

The expression for heat absorbed at B is given by,

  $Q_{\text{B}}=-\frac{3}{2}nR\left(T_{\text{C}}-T_{\text{B}}\right)$

The expression for heat absorbed at C is given by,

  $Q_{\text{C}}=nR{T}_{\text{C}}\ln \frac{V_{\text{D}}}{V_{\text{C}}}$

The expression for total work done is given by,

  $W_{\text{by}\text{tot}}=W_{\text{A}}+W_{\text{B}}+W_{\text{C}}+W_{\text{D}}$

Calculation:

The volume at point D is calculated as,

  $\begin{array}{c}{V}_{\text{D}}=\frac{nR{T}_{\text{D}}}{P_{\text{D}}}\\ =\frac{\left(2\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(360\text{K}\right)}{\left(\left(2.0\text{atm}\right)\left(\frac{101.3\text{kPa}}{1\text{atm}}\right)\right)}\\ =29.5\text{L}\end{array}$

The pressure at point C is calculated as,

  $\begin{array}{c}{P}_{\text{C}}=\frac{nR{T}_{\text{C}}}{V_{\text{C}}}\\ =\frac{\left(2\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(360\text{K}\right)}{\left(3V_{\text{D}}\right)}\\ =\frac{\left(2\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(360\text{K}\right)}{\left(3\left(\left(29.5\text{L}\right)\left(\frac{101.3\text{atm}}{1\text{L}}\right)\right)\right)}\\ =0.667\text{atm}\end{array}$

The temperature at point A and B is calculated as,

  $\begin{array}{c}{T}_{\text{A}}=\frac{P_{\text{A}}{V}_{\text{A}}}{nR}\\ =\frac{\left(2P_{\text{C}}\right)\left(88.6\text{L}\right)}{\left(\text{2}\text{mol}\right)\left(8.206\times {10}^{-2}\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\right)}\\ =\frac{\left(2\left(0.667\text{atm}\right)\right)\left(88.6\text{L}\right)}{\left(\text{2}\text{mol}\right)\left(8.206\times {10}^{-2}\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\right)}\\ =720\text{K}\end{array}$

The heat absorbed at point D is calculated as,

  $\begin{array}{c}{Q}_{\text{D}}=\frac{3}{2}nR\Delta T_{\text{D}}\\ =\frac{3}{2}\left(\text{2}\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}-360\text{K}\right)\\ =\left(\left(8.98\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{kJ}}\right)\right)\\ =8.98\text{kJ}\end{array}$

The heat absorbed at point A is calculated as,

  $\begin{array}{c}{Q}_{\text{A}}=nR{T}_{\text{A}}\ln \frac{V_{\text{B}}}{V_{\text{A}}}\\ =\left(\text{2}\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}\right)\ln \left(\frac{88.6\text{L}}{29.5\text{L}}\right)\\ =\left(\left(13.2\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{kJ}}\right)\right)\\ =13.2\text{kJ}\end{array}$

The heat absorbed at point B is calculated as,

  $\begin{array}{c}{Q}_{\text{B}}=-\frac{3}{2}nR\Delta T_{\text{C}}\\ =-\frac{3}{2}\left(\text{2}\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}-360\text{K}\right)\\ =-\left(\left(8.98\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{kJ}}\right)\right)\\ =-8.98\text{kJ}\end{array}$

The heat absorbed at point C is calculated as,

  $\begin{array}{c}{Q}_{\text{C}}=nR{T}_{\text{C}}\ln \frac{V_{\text{D}}}{V_{\text{C}}}\\ =\left(\text{2}\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(360\text{K}\right)\ln \left(\frac{29.5\text{L}}{88.6\text{L}}\right)\\ =\left(\left(-6.58\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{kJ}}\right)\right)\\ =-6.58\text{kJ}\end{array}$

The total work done is calculated as,

  $\begin{array}{c}{W}_{\text{by}\text{tot}}=W_{\text{A}}+W_{\text{B}}+W_{\text{C}}+W_{\text{D}}\\ =0+13.2\text{kJ}+0-6.58\text{kJ}\\ =6.62\text{kJ}\end{array}$

Conclusion:

Therefore, the total work done is $6.62\text{kJ}$ . The heat absorbed at point A, B, C and D are $13.2\text{kJ}$ , $-8.98\text{kJ}$ , $-6.58\text{kJ}$ and $8.98\text{kJ}$ respectively.