Chapter 21, Problem 65AP

(a)

To determine

The number of moles in the sample.

Answer to Problem 65AP

The number of moles in the sample is $0.203\text{mol}$.

Explanation of Solution

Write the expression from ideal gas law.

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

Rearrange the above equation for $n$.

  $n=\frac{PV}{RT}$                                                                                                      (I)

Conclusion:

Substitute $1.00\text{atm}$ for $P$, $5.00\text{L}$ for $V$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $300\text{K}$ for $T$ in equation (I) to find $n$.

  $\begin{array}{c}n=\frac{\left(1.00\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(5.00\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)}{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(300\text{K}\right)}\\ =\frac{\left(1.013\times {10}^5\text{Pa}\right)\left(5.00\times {10}^{-3}{\text{m}}^3\right)}{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(300\text{K}\right)}\\ =0.203\text{mol}\end{array}$

Therefore, the number of moles in the sample is $0.203\text{mol}$.

(b)

To determine

The temperature of the gas at point B.

Answer to Problem 65AP

The temperature of the gas at point B is $900\text{K}$.

Explanation of Solution

Write the expression for temperature of the sample gas at point B.

  $T_B=T_A\left(\frac{P_B}{P_A}\right)$                                                                                           (II)

Here, $T_B$ is the temperature of the sample gas at point B, $T_A$ is the temperature of the sample gas at point A, $P_B$ is the pressure of the sample gas at point B, and $P_A$ is the pressure of the sample gas at point A.

Conclusion:

Substitute $300\text{K}$ for $T_A$, $3.00\text{atm}$ for $P_B$, and $1.00\text{atm}$ for $P_A$ in equation (II) to find $T_B$.

  $\begin{array}{c}{T}_B=\left(300\text{K}\right)\left(\frac{3.00\text{atm}}{1.00\text{atm}}\right)\\ =900\text{K}\end{array}$

Therefore, the temperature of the gas at point B is $900\text{K}$.

(c)

To determine

The temperature of the gas at point C.

Answer to Problem 65AP

The temperature of the gas at point C is $900\text{K}$.

Explanation of Solution

The temperature of the gas at point C is equal to the temperature at point B.

  $T_C=T_B$

Conclusion:

Substitute $900\text{K}$ for $T_B$ in the above equation.

  $T_C=900\text{K}$

Therefore, the temperature of the gas at point C is $900\text{K}$.

(d)

To determine

The volume at the point C.

Answer to Problem 65AP

The volume at the point C is $15.0\text{L}$.

Explanation of Solution

The figure 1 shows the PV diagram.

Write the expression for volume of the sample gas at point C.

  $V_C=V_A\left(\frac{T_C}{T_A}\right)$                                                                                           (III)

Here, $T_C$ is the temperature of the sample gas at point C, $T_A$ is the temperature of the sample gas at point A, $V_C$ is the volume of the sample gas at point C, and $V_A$ is the volume of the sample gas at point A.

Conclusion:

Substitute $300\text{K}$ for $T_A$, $900\text{K}$ for $T_C$, and $5.00\text{L}$ for $V_A$ in equation (III) to find $V_C$.

  $\begin{array}{c}{V}_C=\left(5.00\text{L}\right)\left(\frac{900\text{K}}{300\text{K}}\right)\\ =15.0\text{L}\end{array}$

Therefore, the volume at the point C is $15.0\text{L}$.

(e)

To determine

How to carry out the process $A\to B$, $B\to C$, and $C\to A$ experimentally.

Answer to Problem 65AP

The process $A\to B$ is to lock the piston $B\to C$ is keep the sample in oven, and $C\to A$ is carry the cylinder experimentally.

Explanation of Solution

$A\to B$ process carried out experimentally by lock the piston in place and put the cylinder inot an oven at $900\text{K}$, gradually by heating the gas.

$B\to C$ process keep the sample in the oven while gradually letting the gas expand to lift a load on the piston.

Conclusion:

$C\to A$ process carried out experimentally by carry the cylinder back into the room at $300\text{K}$ and let the gas gradually cool and contract without touching the piston.

(f)

To determine

Find $Q,W,\text{ and }\Delta E_{\mathrm{int}}$ for each process.

Answer to Problem 65AP

The $A\to B$ process $Q$ value is $1.52\text{kJ}$, $W$ value is $0$ and $\Delta E_{\mathrm{int}}$ is $1.52\text{kJ}$, $B\to C$ process $Q$ value is $1.67\text{kJ}$, $W$ value is $-1.67\text{kJ}$ and $\Delta E_{\mathrm{int}}$ is $0$, $C\to A$ process $Q$ value is $-2.53\text{kJ}$, $W$ value is $1.01\text{kJ}$, and $\Delta E_{\mathrm{int}}$ is $-1.52\text{kJ}$.

Explanation of Solution

For $A\to B$ process $W=0$.

Find the internal energy of the gas.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\Delta T\end{array}$ (IV)

Here, $C_V$ is the specific heat capacity at constant volume and $\Delta T$ is the change in temperature

Find the heat energy for the given process.

  $Q=\Delta E_{\mathrm{int}}-W$                                                                                            (V)

For $B\to C$ process $\Delta E_{\mathrm{int}}=0$ because $\Delta T=0$.

Find the work done on the gas.

  $W=-nR{T}_B\ln \left(\frac{V_C}{V_B}\right)$                                                                                  (VI)

Find the heat energy for the given process.

  $Q=\Delta E_{\mathrm{int}}-W$                                                                                          (VII)

For $C\to A$ process,

Find the internal energy of the gas.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\Delta T\end{array}$                                                                                 (VIII)

Find the work done on the gas.

  $\begin{array}{c}W=-P\Delta V\\ =-nR\Delta T\end{array}$                                                                                           (IX)

Find the heat energy for the given process.

  $Q=\Delta E_{\mathrm{int}}-W$                                                                                         (X)

Conclusion:

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $900\text{K}$ for $\Delta T$ in equation (IV) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{3}{2}\left(0.203\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(600\text{K}\right)\\ =1.52\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =1.52\text{kJ}\end{array}$

Substitute $1.52\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $0$ for $W$ in equation (V) to find $Q$.

  $\begin{array}{c}Q=1.52\text{kJ}-0\\ =1.52\text{kJ}\end{array}$

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $3.00$ for $V_C/V_B$, and $900\text{K}$ for $\Delta T$ in equation (VI) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}W=-\left(0.203\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(900\text{K}\right)\ln \left(3.00\right)\\ =-1.67\text{kJ}\end{array}$

Substitute $0$ for $\Delta E_{\mathrm{int}}$ and $-1.67\text{kJ}$ for $W$ in equation (VII) to find $Q$.

  $\begin{array}{c}Q=0-\left(-1.67\text{kJ}\right)\\ =1.67\text{kJ}\end{array}$

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $-600\text{K}$ for $\Delta T$ in equation (VIII) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{3}{2}\left(0.203\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(-600\text{K}\right)\\ =-1.52\text{kJ}\end{array}$

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $-600\text{K}$ for $\Delta T$ in equation (IX) to find $W$.

  $\begin{array}{c}W=-\left(0.203\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(-600\text{K}\right)\\ =1.01\text{kJ}\end{array}$

Substitute $-1.52\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $1.01\text{kJ}$ for $W$ in equation (X) to find $Q$.

  $\begin{array}{c}Q=-1.52\text{kJ}-1.01\text{kJ}\\ =-2.53\text{kJ}\end{array}$

Therefore, the $A\to B$ process $Q$ value is $1.52\text{kJ}$, $W$ value is $0$ and $\Delta E_{\mathrm{int}}$ is $1.52\text{kJ}$, $B\to C$ process $Q$ value is $1.67\text{kJ}$, $W$ value is $-1.67\text{kJ}$ and $\Delta E_{\mathrm{int}}$ is $0$, $C\to A$ process $Q$ value is $-2.53\text{kJ}$, $W$ value is $1.01\text{kJ}$, and $\Delta E_{\mathrm{int}}$ is $-1.52\text{kJ}$.

(g)

To determine

Find $Q_{ABCA},W_{ABCA},\text{ and }{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}$ for whole cycle process.

Answer to Problem 65AP

The whole cycle process $Q_{ABCA}$ value is $0.656\text{kJ}$, $W_{ABCA}$ value is $-0.656\text{kJ}$ and ${\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}$ is $0$.

Explanation of Solution

Find the heat energy for the whole cycle process.

  $Q_{ABCA}=Q_{A\to B}+Q_{B\to C}+Q_{C\to A}$                                                                         (XI)

Find the work done on the gas for the whole cycle process.

  $W_{ABCA}=W_{A\to B}+W_{B\to C}+W_{C\to A}$                                                                      (XII)

Find the internal energy of the gas for the whole cycle process.

  ${\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=Q_{ABCA}+W_{ABCA}$                                                                             (XIII)

Conclusion:

Substitute $1.52\text{kJ}$ for $Q_{A\to B}$, $1.67\text{kJ}$ for $Q_{B\to C}$, and $-2.53\text{kJ}$ for $Q_{C\to A}$ in equation (XI) to find $Q_{ABCA}$.

  $\begin{array}{c}{Q}_{ABCA}=1.52\text{kJ}+1.67\text{kJ}-2.53\text{kJ}\\ =0.656\text{kJ}\end{array}$

Substitute $0$ for $W_{A\to B}$, $-1.67\text{kJ}$ for $W_{B\to C}$, and $1.01\text{kJ}$ for $W_{C\to A}$ in equation (XII) to find $W_{ABCA}$.

  $\begin{array}{c}{W}_{ABCA}=0-1.67\text{kJ+}1.01\text{kJ}\\ =-0.656\text{kJ}\end{array}$

Substitute $0.656\text{kJ}$ for $Q_{ABCA}$ and $-0.656\text{kJ}$ for $W_{ABCA}$ in equation (XIII) to find ${\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}$.

  $\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=0.656\text{kJ}-0.656\text{kJ}\\ =\text{0}\end{array}$

Therefore, the whole cycle process $Q_{ABCA}$ value is $0.656\text{kJ}$, $W_{ABCA}$ value is $-0.656\text{kJ}$ and ${\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}$ is $0$.