Chapter 15, Problem 20P

(a)

To determine

The total work done on the gas if it follows the constant temperature path $AC$ followed by the constant–pressure path $CD$.

Answer to Problem 20P

The total work done on the gas is $\underset{\_}{-1.37\text{kJ}}$.

Explanation of Solution

Write the expression for the total work done when the gas follows the constant temperature path followed by constant pressure path.

    $W_{\text{total}}=W_{\text{t}}+W_p$                                                                                                         (I)

Here, $W_{\text{total}}$ is the total work done, $W_t$ is the work done by constant temperature, $W_v$ is the work done by constant volume.

it can be shown from the $PV$ curve that the work done on an ideal gas during constant temperature expansion or contraction from volume $v_I$ to volume $v_F$ can be represented as,

    $W_{\text{t}}=nRT\ln \frac{V_i}{V_f}$                                                                                                     (II)

Here, $W_{\text{total}}$ is the work done by the gas under constant temperature, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V_i$ is the initial volume, $V_f$ is the final volume.

Write the expression for ideal gas equation.

    $PV=nRT$                                                                                                             (III)

Here, $P$ is the pressure of the gas, $V$ is the volume of the gas.

Solve equation (III) for $T$,

    $T=\frac{PV}{nR}$                                                                                                                 (IV)

Use equation (IV) in equation (II),

    $\begin{array}{c}{W}_t=nR\left(\frac{PV}{nR}\right)\ln \frac{V_i}{V_f}\\ =PV\ln \frac{V_i}{V_f}\end{array}$                                                                                                (V)

Write the expression for work done by the gas at constant pressure.

    $W_p=-P_i\Delta V$                                                                                                           (VI)

Here, $W_p$ is the work done at constant pressure, $\Delta V$ is the change in volume.

Use equation (IV) and (III) in equation (I),

    $W_{\text{total}}=PV\ln \frac{V_i}{V_f}-P_i\Delta V$                                                                                     (VII)

Rearrange equation (V) for work done when the gas follows the constant temperature path $AC$ followed by the constant pressure path $CD$.

    $W_{\text{total}}=P_A{V}_A\ln \frac{V_A}{V_C}-P_i\Delta V$                                                                                    (VIII)

Conclusion:

Substitute $2.00\text{atm}$ for $P_A$, $4.00\text{L}$ for $V_A$, $8.00\text{L}$ for $V_C$, $1.00\text{atm}$ for $P_I$,$8.00\text{L}$ for $\Delta V$ in equation (VIII) to find $W_{\text{total}}$.

    $\begin{array}{c}{W}_{\text{total}}=\left(2.00\text{atm}\right)\left(4.00\text{L}\right)\ln \left(\frac{4.00\text{L}}{8.00\text{L}}\right)-\left(1.00\text{atm}\right)\left(8.00\text{L}\right)\\ =\left[\left(2.00\text{atm}\right)\left(4.00\text{L}\right)\ln \left(\frac{4.00\text{L}}{8.00\text{L}}\right)-\left(1.00\text{atm}\right)\left(8.00\text{L}\right)\right]\left(1.013\times {10}^5\text{Pa}/\text{atm}\right)\times \frac{1{\text{m}}^3}{1000\text{L}}\\ =1.37\text{kJ}\end{array}$

Therefore, the total work done on the gas is $\underset{\_}{-1.37\text{kJ}}$.

(b)

To determine

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

Answer to Problem 20P

The total change in internal energy is $\underset{\_}{1.22\text{kJ}}$ and total heat flow into the gas is $\underset{\_}{2.59\text{kJ}}$.

Explanation of Solution

Write the expression for the change in internal energy.

    $\Delta U=n{C}_{\text{V}}\Delta T$                                                                                                         (IX)

Here, $\Delta U$ is the change in internal energy, $C_{\text{V}}$ is the specific heat at constant volume, $\Delta T$ is the change in temperature.

Write the expression for the total heat flows into the gas.

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

Write the expression for the specific heat at constant volume.

    $C_{\text{V}}=\frac{3}{2}R$                                                                                                               (XI)

Use equation (X) and equation (IV) in equation (IX),

    $\begin{array}{c}\Delta U=\frac{3}{2}nR\left(\frac{P\Delta V}{nR}\right)\\ =\frac{3}{2}P\Delta V\end{array}$                                                                                             (XII)

Conclusion:

Substitute $1.00\text{atm}$ for $P$, $8.00\text{L}$ for $\Delta V$ in equation (XII) to find $\Delta U$.

    $\begin{array}{c}\Delta U=\frac{3}{2}\left(1.00\text{atm}\right)\left(8.00\text{L}\right)\\ =\left[\frac{3}{2}\left(1.00\text{atm}\right)\left(8.00\text{L}\right)\right]\times \left(1.013\times {10}^5\text{Pa}/\text{atm}\right)\times \frac{1{\text{m}}^3}{1000\text{L}}\\ =1.22\text{kJ}\end{array}$

Substitute $-1.37\text{kJ}$ for $W$ and $1.22\text{kJ}$ for $\Delta U$ in equation (X) to find $Q$.

    $\begin{array}{c}Q=1.22\text{kJ}-\left(-1.37\text{kJ}\right)\\ =2.59\text{kJ}\end{array}$

Therefore, The total change in internal energy is $\underset{\_}{1.22\text{kJ}}$ and total heat flow into the gas is $\underset{\_}{2.59\text{kJ}}$.