Chapter 18, Problem 69P

(a)

To determine

The final temperature, volume, work done and heat absorbed if the expansion is isothermal.

Answer to Problem 69P

The final temperature, volume, work done and heat absorbed are $300\text{K}$ , $7.80\text{L}$ , $1.14\text{kJ}$ and $1.14\text{kJ}$ respectively.

Explanation of Solution

Given:

The initial pressure is $400\text{kPa}$ .

The final pressure is $\text{160}\text{kPa}$

The initial temperature is $300\text{K}$ .

Formula used:

The expression for initial volume is given by,

  $V_{\text{i}}=\frac{nRT}{P}$

The expression for final volume is given by,

  $V_{\text{f}}=V_{\text{i}}\frac{P_{\text{i}}}{P_{\text{f}}}$

The expression for work done is given by,

  $W_{\text{by}\text{gas}}=nRT\ln \frac{V_{\text{f}}}{V_{\text{i}}}$

The expression for heat absorbed is given by,

  $Q_{\text{in}}=\Delta E_{\mathrm{int}}+W_{\text{bygas}}$

Calculation:

The temperature remains same for an isothermal expansion.

The initial volume is calculated as,

  $\begin{array}{c}{V}_{\text{i}}=\frac{nRT}{P}\\ =\frac{\left(0.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)}{400\text{kPa}}\\ =\left(\left(3.12\times {10}^{-3}{\text{m}}^3\right)\left(\frac{{10}^3\text{L}}{1{\text{m}}^3}\right)\right)\\ =3.12\text{L}\end{array}$

The final volume is calculated as,

  $\begin{array}{c}{V}_{\text{f}}=V_{\text{i}}\frac{P_{\text{i}}}{P_{\text{f}}}\\ =\left(3.12\text{L}\right)\left(\frac{400\text{kPa}}{160\text{kPa}}\right)\\ =7.80\text{L}\end{array}$

The work done by gas is calculated as,

  $\begin{array}{c}{W}_{\text{by}\text{gas}}=nRT\ln \frac{V_{\text{f}}}{V_{\text{i}}}\\ =\left(0.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)\ln \left(\frac{7.80\text{L}}{3.12\text{L}}\right)\\ =\left(\left(1.14\times {10}^3\text{J}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\right)\\ =1.14\text{kJ}\end{array}$

The heat absorbed is calculated as,

  $\begin{array}{c}{Q}_{\text{in}}=\Delta E_{\mathrm{int}}+W_{\text{bygas}}\\ =0+\left(1.14\text{kJ}\right)\\ =1.14\text{kJ}\end{array}$

Conclusion:

Therefore, the final temperature, volume, work done and heat absorbed are $300\text{K}$ , $7.80\text{L}$ , $1.14\text{kJ}$ and $1.14\text{kJ}$ respectively.

(b)

To determine

The final temperature, volume, work done and heat absorbed if the expansion is adiabatic.

Answer to Problem 69P

The final temperature, volume, work done and heat absorbed are $208\text{K}$ , $5.41\text{L}$ , $574\text{J}$ and $0$ respectively.

Explanation of Solution

Formula used:

The expression for final temperature is given by,

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

The expression for final volume is given by,

  $V_{\text{f}}=V_{\text{i}}{\left(\frac{P_{\text{i}}}{P_{\text{f}}}\right)}^r$

The expression for work done is given by,

  $W=\frac{3}{2}nR\Delta T$

Calculation:

The final volume is calculated as,

  $\begin{array}{c}{V}_{\text{f}}=V_{\text{i}}{\left(\frac{P_{\text{i}}}{P_{\text{f}}}\right)}^r\\ =\left(3.12\text{L}\right){\left(\frac{400\text{kPa}}{160\text{kPa}}\right)}^{\frac{3}{5}}\\ =5.41\text{L}\end{array}$

The final temperature is calculated as,

  $\begin{array}{l}{T}_{\text{f}}=\frac{P_{\text{f}}{V}_{\text{f}}}{nR}\\ =\frac{\left(160\text{kPa}\right)\left(\left(5.41\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)}{\left(0.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =208\text{K}\end{array}$

The work done by gas is calculated as,

  $\begin{array}{c}W=\frac{3}{2}nR\Delta T\\ =\frac{3}{2}\left(0.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}-280\text{K}\right)\\ =574\text{J}\end{array}$

The heat absorbed is zero in case of adiabatic process.

Conclusion:

Therefore, the final temperature, volume, work done and heat absorbed are $208\text{K}$ , $5.41\text{L}$ , $574\text{J}$ and $0$ respectively.