Chapter 22, Problem 42PQ

(a)

To determine

The change in entropy of the cold reservoir.

Answer to Problem 42PQ

The change in entropy of the cold reservoir is $64.2\text{J}/\text{K}$.

Explanation of Solution

Write the expression for the change in entropy.

    $\Delta S=\frac{\Delta Q}{T}$

Here, $\Delta S$ is the change in entropy, $\Delta Q$ is the energy transferred, and $T$ is the temperature.

Conclusion:

Substitute $2.35\text{atm}$ for $P$, $0.340{\text{m}}^{\text{3}}$ for $V_f$ , and $0.525{\text{m}}^{\text{3}}$ for $V_i$ to find $W$.

    $\begin{array}{c}W=-\left(2.35\text{atm}\right)\left(0.340{\text{m}}^{\text{3}}-0.525{\text{m}}^{\text{3}}\right)\\ =-\left(2.35\text{atm}\right)\left(\frac{1.01\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(0.340{\text{m}}^{\text{3}}-0.525{\text{m}}^{\text{3}}\right)\\ =4.40\times {10}^4\text{J}\end{array}$

Therefore, the work done on the gas is $4.40\times {10}^4\text{J}$.

(b)

To determine

The heat transferred when a monoatomic carbon gas undergoes a constant pressure process.

Answer to Problem 42PQ

The heat transferred when a monoatomic carbon gas undergoes a constant pressure process

is $-1.10\times {10}^5\text{J}$

Explanation of Solution

Write the expression initial temperature from ideal gas equation.

    $T_i=\frac{P{V}_i}{nR}$                                                                                                                      (I)

Here, $T_i$ is the initial temperature, $n$ is the number of moles, and $R$ is the ideal gas constant.

Write the expression final temperature from ideal gas equation.

    $T_f=\frac{P{V}_f}{nR}$                                                                                                                   (II)

Here, $T_f$ is the final temperature.

Write the expression for the heat transfer.

    $Q_P=n{C}_P\left(T_f-T_i\right)$                                                                                                   (III)

Here, $Q_P$ is the amount of heat transfer and $C_P$ is the specific heat of the monoatomic gas at constant pressure.

For monoatomic gas,

    $C_P=\frac{5}{2}R$                                                                                                                (IV)

Use (I), (II),(IV) to rewrite (III).

    $\begin{array}{c}{Q}_P=n\left(\frac{5}{2}R\right)\left(\frac{P{V}_f}{nR}-\frac{P{V}_i}{nR}\right)\\ =\frac{5}{2}P\left(V_f-V_i\right)\end{array}$                                                                                   (V)

Conclusion:

Substitute $2.35\text{atm}$ for $P$, $0.340{\text{m}}^{\text{3}}$ for $V_f$ , and $0.525{\text{m}}^{\text{3}}$ for $V_i$ to find $W$.

    $\begin{array}{c}W=\left(\frac{5}{2}\right)\left(2.35\text{atm}\right)\left(0.340{\text{m}}^{\text{3}}-0.525{\text{m}}^{\text{3}}\right)\\ =\left(\frac{5}{2}\right)\left(2.35\text{atm}\right)\left(\frac{1.01\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(0.340{\text{m}}^{\text{3}}-0.525{\text{m}}^{\text{3}}\right)\\ =-1.10\times {10}^5\text{J}\end{array}$

Therefore, the heat transferred when a monoatomic carbon gas undergoes a constant pressure process is $-1.10\times {10}^5\text{J}$.

(c)

To determine

The change in entropy.

Answer to Problem 42PQ

The change in entropy is $-271\text{J}/\text{K}$

Explanation of Solution

Write the expression for change in entropy.

    $\begin{array}{c}\Delta S={\displaystyle {\int }_i^f\frac{n\left(\frac{5}{2}R\right)dT}{T}}\\ =n\left(\frac{5}{2}R\right){\displaystyle {\int }_i^f\frac{dT}{T}}\\ =n\left(\frac{5}{2}R\right){\left[\ln \left(T\right)\right]}_i^f\\ =n\left(\frac{5}{2}R\right)\ln \left(T_f\right)-\ln \left(T_i\right)\\ =n\left(\frac{5}{2}R\right)\ln \left(\frac{T_f}{T_i}\right)\end{array}$                                                                                  (VI)

Here, $\Delta S$ is the change in entropy.

Use (I) and (II) to rewrite (VI).

    $\begin{array}{c}\Delta S=n\left(\frac{5}{2}R\right)\ln \left(\frac{\frac{P{V}_f}{nR}}{\frac{P{V}_i}{nR}}\right)\\ =n\left(\frac{5}{2}R\right)\ln \left(\frac{V_f}{V_i}\right)\end{array}$                                                                                        (VII)

Conclusion:

Substitute $30.0\text{mol}$ for $n$ , $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for $R$ , $0.340{\text{m}}^{\text{3}}$ for $V_f$ , and $0.525{\text{m}}^{\text{3}}$ for $V_i$  in (VII) to find $Q_P$.

    $\begin{array}{c}{Q}_P=\left(30.0\text{mol}\right)\left(\frac{5}{2}\right)\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\ln \left(\frac{0.340{\text{m}}^{\text{3}}}{0.525{\text{m}}^{\text{3}}}\right)\\ =-270.9\text{J}/\text{K}\\ \approx -271\text{J}/\text{K}\end{array}$

Therefore, the coefficient of performance of an ideal heat pump is $-271\text{J}/\text{K}$