Chapter 17, Problem 68P

A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig. P17.68). It is warmed at constant volume to 3.00 atm (point B). Then it is allowed to expand isothermally to 1.00 atm (point C) and at last compressed isobarically to its original state. (a) Find the number of moles in the sample. Find (b) the temperature at point B, (c) the temperature at point C, and (d) the volume at point C. (e) Now consider the processes A → B, B → C, and C → A. Describe how to carry out each process experimentally. (f) Find Q, W, and ΔE_int for each of the processes. (g) For the whole cycle A → B → C → A, find Q, W, and ΔE_int.

Figure P17.68

To determine

Thenumber of mole in the sample.

Answer to Problem 68P

Thenumber of mole in the sample is $\underset{\_}{0.203\text{mol}}$.

Explanation of Solution

Write the expression for theideal gas law,

  $PV=nRT$        (I)

Here, $P$ is the pressure of the sample, $V$ is the volume of the sample, $n$  is the mole in the sample, $R$ is the universal gas constant and $T$ is the temperature of the sample.

Rewrite the above equation,

    $n=\frac{RT}{PV}$        (II)

Conclusion:

Substitute $1.013\times {10}^5\text{Pa}$ for $P$, $5.00\times {10}^{-3}{\text{m}}^{\text{3}}$ for $V$, $8.314\text{J/mol}\text{.K}$ for $R$ and $300\text{K}$ for $T$ in (II) to find $n$ ,

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

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

To determine

The temperature of the sample at point $B$.

Answer to Problem 68P

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

Explanation of Solution

Write the expression for thetemperature of the sample at point $B$,

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

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

Conclusion:

Substitute $300\text{K}$ for $T_A$, $3.00$ for $P_B$ and $1.00$ for $P_A$ in (III),

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

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

To determine

The temperature of the sample at point $C$.

Answer to Problem 68P

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

Explanation of Solution

Write the expression for thetemperature of the sample at point $C$,

    $T_C=T_B$        (IV)

Here, $T_C$ is the temperature of the sample at point $C$ .

Conclusion:

Substitute $900\text{K}$ for $T_C$, in (IV),

    $T_C=900\text{K}$

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

To determine

The volume of the sample at point $C$.

Answer to Problem 68P

The volume of the sample at point $C$ is $\underset{\_}{15.0\text{L}}$ .

Explanation of Solution

Write the expression for thevolume of the sample at point $C$,

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

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

Conclusion:

Substitute $5.00\text{L}$ for $V_A$, $900\text{K}$ for $T_C$ and $300\text{K}$ for $T_A$ in (V),

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

Therefore, the volume of the sample at point $C$ is $\underset{\_}{15.0\text{L}}$

To determine

Explain the processes $A\to B,B\to C\text{and}C\to A$.

Answer to Problem 68P

In the process $A\to B$ lock the piston in place and put the cylinder into an oven.

In the process $B\to A$ the gas expand to lift a load on the piston.

In the process $C\to A$ gas gradually cool and contract without touching the piston.

Explanation of Solution

In the process $A\to B$,

Lock the piston in place and put the cylinder into an oven at the temperature of $900\text{K}$ .

Here the gas is getting heat gradually.

In the process $B\to A$,

The sample is keep in the oven while the gas expand gradually and to lift a load on the piston as far as it can.

In the process $C\to A$,

The cylinder is carry and back into the room temperature at $300\text{K}$.

Here the gas is cooling gradually and contract without touching the piston.

Conclusion:

Therefore, in the process $A\to B$ lock the piston in place and put the cylinder into an oven.

In the process $B\to A$ the gas expand to lift a load on the piston.

In the process $C\to A$ gas gradually cool and contract without touching the piston.

To determine

The value of $Q,Wand\Delta E_{\mathrm{int}}$ for each process .

Answer to Problem 68P

The value of $Q,Wand\Delta E_{\mathrm{int}}$ for each process are $\underset{\_}{A\to B:W=0,\Delta E_{\mathrm{int}}=1.52\text{kJ,}Q=1.52\text{kJ}}$, $\underset{\_}{B\to C:W=-1.67\text{kJ},\Delta E_{\mathrm{int}}=0\text{,}Q=1.67\text{kJ}}$ and $\underset{\_}{C\to A:W=1.01\text{kJ},\Delta E_{\mathrm{int}}=-1.52\text{kJ,}Q=-2.53\text{kJ}}$.

Explanation of Solution

For $A\to B$,

Write the expression for theenergy transferred by heat,

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

Here, $Q$ is the energy transferred by heat, $\Delta E_{\mathrm{int}}$ is the change in internal energy and $W$ is the work done on the sample.

Write the expression for the change in internal energy,

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

In this case, $W=0$.

For $B\to A$,

Write the expression for the work done on the sample,

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

In this case, $\Delta E_{\mathrm{int}}=0$.

For $C\to A$,

Write the expression for the work done on the sample,

    $W=-nR\Delta T$        (IX)

Conclusion:

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\text{.K}$ for $R$ and $600\text{K}$ for $\Delta T$ in (VI),

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

Substitute $1.52\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $0$ for $W$ in (VI),

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

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\text{.K}$ for $R$, $900\text{K}$ for $T_B$ and $3.00$ for $\frac{V_C}{V_B}$ in (VIII),

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

Substitute $-1.67\text{kJ}$ for $W$ and $0$ for $\Delta E_{\mathrm{int}}$ in (VI),

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

Substitute $0.203\text{mol}$ for $n$, $8.314\text{J/mol}\text{.K}$ for $R$ and $-600\text{K}$ for $\Delta T$ in (VII),

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

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

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

Substitute $-1.52\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $1.01\text{kJ}$ for $W$ in (VI),

    $Q=-1.52\text{kJ}-1.01\text{kJ}$

Therefore, the value of $Q,Wand\Delta E_{\mathrm{int}}$ for each process are $\underset{\_}{A\to B:W=0,\Delta E_{\mathrm{int}}=1.52\text{kJ,}Q=1.52\text{kJ}}$, $\underset{\_}{B\to C:W=-1.67\text{kJ},\Delta E_{\mathrm{int}}=0\text{,}Q=1.67\text{kJ}}$ and $\underset{\_}{C\to A:W=1.01\text{kJ},\Delta E_{\mathrm{int}}=-1.52\text{kJ,}Q=-2.53\text{kJ}}$.

To determine

The value of $Q,Wand\Delta E_{\mathrm{int}}$ for whole process .

Answer to Problem 68P

The values of $Q,Wand\Delta E_{\mathrm{int}}$ for whole process are

$\underset{\_}{Q_{ABCA}=0.656\text{kJ, }{W}_{ABCA}=-0.656\text{kJ}\text{and}{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=0}$.

Explanation of Solution

Write the expression for theenergy transferred by heat in the whole process

    $Q_{ABCA}=Q_{AB}+Q_{BC}+Q_{CA}$        (X)

Write the expression for the work done for whole process,

    $W_{ABCA}=W_{AB}+W_{BC}+W_{CA}$        (XI)

Write the expression for the change in internal energy for whole process,

    ${\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}={\left(\Delta E_{\mathrm{int}}\right)}_{AB}+{\left(\Delta E_{\mathrm{int}}\right)}_{BC}+{\left(\Delta E_{\mathrm{int}}\right)}_{CA}$        (XII)

Conclusion:

Substitute $1.52\text{kJ}$ for $Q_{AB}$, $1.67\text{kJ}$ for $Q_{BC}$ and $-2.53\text{kJ}$ for $Q_{CA}$ in (X),

    $\begin{array}{c}{W}_{ABCA}=\left(1.52\text{kJ}\right)+\left(1.67\text{kJ}\right)+\left(-2.53\text{kJ}\right)\\ =0.656\text{kJ}\end{array}$

Substitute $0$ for $W_{AB}$, $-1.67\text{kJ}$ for $W_{BC}$ and $1.01\text{kJ}$ for $W_{CA}$ in (XI),

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

Substitute $1.52\text{kJ}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{AB}$, $0$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{BC}$ and $-1.52\text{kJ}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{CA}$ in (XII),

    $\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=1.52\text{kJ}+0-1.52\text{kJ}\\ =0\end{array}$

Therefore, the values of $Q,Wand\Delta E_{\mathrm{int}}$ for whole process are $\underset{\_}{Q_{ABCA}=0.656\text{kJ, }{W}_{ABCA}=-0.656\text{kJ}\text{and}{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=0}$