Chapter 15, Problem 21P

(a)

To determine

The work done by the gas if it follows a constant pressure path $AE$, and constant temperature path $ED$.

Answer to Problem 21P

The work done by the gas is $\underset{\_}{-1.93\text{kJ}}$.

Explanation of Solution

Write the expression for work done at constant pressure $AE$.

$W=-P\Delta V_{\text{i}}$ (I)

Here, $W$ is the work done, $P$ is the pressure, $\Delta V$ is the change in volume.

Write the expression for work done at constant temperature $ED$.

$W=nRT\ln \frac{V_{\text{E}}}{V_{\text{D}}}$ (II)

Here, $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the temperature, $V_{\text{E}}$ is the volume at E, and $V_D$ is the volume at $D$.

Write the expression for work done for a combined constant pressure and temperature process.

$W_{\text{total}}=-P\Delta V+nRT\ln \frac{V_{\text{E}}}{V_{\text{D}}}$ (III)

Write the ideal gas equation to obtain an expression for $T$.

$\begin{array}{c}{P}_{\text{E}}{V}_{\text{E}}=nRT\\ T=\frac{P_{\text{E}}{V}_{\text{E}}}{nR}\end{array}$ (IV)

Substitute equation (IV) in (III).

$W_{\text{total}}=-P\Delta V+nR\left(\frac{P_{\text{E}}{V}_{\text{E}}}{nR}\right)\ln \frac{V_{\text{E}}}{V_{\text{D}}}$ (V)

Conclusion:

Substitute $2.0\text{atm}$ for $P$, $\left(8.0\text{L-4}\text{.0}\text{L}\right)$ for $\Delta V$, $8.0\text{L}$ for $V_{\text{E}}$, $16.0\text{L}$ for $V_D$, in equation (V).

$\begin{array}{c}{W}_{\text{total}}=\left[-\left(2.0\text{atm}\right)\left(\left(8.0\text{L-4}\text{.0}\text{L}\right)\right)+\left(2.0\text{atm}\times 8.0\text{L}\right)\ln \frac{8.0\text{L}}{16.0\text{L}}\right]\left(1.013\times {10}^5\text{Pa/atm}\right)\left({10}^{-3}{\text{m}}^3/\text{L}\right)\\ =-1.933\text{kJ}\end{array}$

Therefore, the work done by the gas is $\underset{\_}{-1.93\text{kJ}}$.

(b)

To determine

The total change in internal energy during entire process and total heat flow into the gas.

Answer to Problem 21P

The total change in internal energy during entire process $\underset{\_}{1.22\text{kJ}}$, and total heat flow into the gas is $\underset{\_}{3.15\text{kJ}}$.

Explanation of Solution

Write the expression for the internal energy change at constant temperature.

$\Delta U=0$ (VI)

Write the expression for the internal energy change at constant pressure.

$\begin{array}{c}\Delta U=n{C}_V\Delta T\\ =\frac{3}{2}nR\Delta T\end{array}$ . (VII)

Here, $C_V$ is the heat capacity at constant volume, and $\Delta T$ is the change in temperature, and $R$ is the universal gas constant.

Write the expression for $\Delta T$ according to equation (VI).

$\begin{array}{c}P\Delta V_{\text{i}}=nR\Delta T\\ \Delta T=\frac{P\Delta V_{\text{i}}}{nR}\end{array}$ (VIII)

Substitute equation (VIII) in (VII).

$\Delta U=\frac{3}{2}nR\left(\frac{P\Delta V_{\text{i}}}{nR}\right)$ . (IX)

Write the expression for the first law of thermodynamics.

$Q=\Delta U-W$ (X)

Conclusion:

Substitute $2.0\text{atm}$ for $P$, $\left(8.0\text{L-4}\text{.0}\text{L}\right)$ for $\Delta V$ in equation (IX).

$\begin{array}{c}\Delta U=\left[\frac{3}{2}\left(2.0\text{atm}\times \left(8.0\text{L-4}\text{.0}\text{L}\right)\right)\right]\left(1.013\times {10}^{-5}\text{Pa/atm}\times {\text{10}}^{-3}{\text{m}}^3/\text{L}\right)\\ =1.22\text{kJ}\end{array}$

Substitute $1.22\text{kJ}$ for $\Delta U$, $-1.933\text{kJ}$ for $W$ in equation (X).

$\begin{array}{c}Q=1.22\text{kJ}-\left(-1.933\text{kJ}\right)\\ =3.15\text{kJ}\end{array}$

Therefore, the total change in internal energy during entire process $\underset{\_}{1.22\text{kJ}}$, and total heat flow into the gas is $\underset{\_}{3.15\text{kJ}}$.