Chapter 21, Problem 21.65AP

A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig. P21.65). 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.

To determine

The number of moles in the sample.

Answer to Problem 21.65AP

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

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

The number of moles in the ideal gas equation is,

$n=\frac{P_{\text{A}}{V}_{\text{A}}}{R{T}_{\text{A}}}$

Here,

$P_{\text{A}}$ is the pressure at point $\text{A}$.

$V_{\text{A}}$ is the volume at point $\text{A}$.

$R$ is the ideal gas constant.

$T_{\text{A}}$ is the temperature at point $\text{A}$.

The value of the ideal gas constant is $8.314\text{J}/\text{mol}\text{.K}$.

Substitute $1\text{atm}$ for $P_{\text{A}}$, $5.00\text{L}$ for $V_{\text{A}}$, $300\text{K}$ for $T_{\text{A}}$ and $8.314\text{J}/\text{mol}\text{.K}$ for $R$ in above equation.

$\begin{array}{c}n=\frac{1\text{atm}\left(\frac{101\times {10}^3\text{Pa}}{1\text{atm}}\right)\times 5.00\text{L}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)}{8.314\text{J}/\text{mol}\text{.K}\times 300\text{K}}\\ =0.203\text{mol}\end{array}$

Conclusion:

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

To determine

The temperature at point $\text{B}$.

Answer to Problem 21.65AP

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

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

In the process from point $\text{A}$ to point $\text{B}$, the volume does not changes.

The expression for the process from point $\text{A}$ to point $\text{B}$ is,

$\frac{P_{\text{A}}}{T_{\text{A}}}=\frac{P_{\text{B}}}{T_{\text{B}}}$

Here,

$P_{\text{B}}$ is the pressure at point $\text{B}$.

$T_{\text{B}}$ is the temperature at point $\text{B}$.

Substitute $1\text{atm}$ for $P_{\text{A}}$, $300\text{K}$ for $T_{\text{A}}$, and $3\text{atm}$ for $P_{\text{B}}$ in above equation.

$\begin{array}{c}\frac{1\text{atm}}{300\text{K}}=\frac{3\text{atm}}{T_{\text{B}}}\\ T_{\text{B}}=900\text{K}\end{array}$

Conclusion:

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

To determine

The temperature at point $\text{C}$.

Answer to Problem 21.65AP

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

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

In the process from point $\text{B}$ to point $\text{C}$, the temperature does not changes because the gas expand isothermally from point $\text{B}$ to point $\text{C}$.

So, the temperature at point $\text{C}$ will be same as temperature at point at point $\text{B}$ which is $900\text{K}$.

Conclusion:

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

To determine

The volume at point $\text{C}$.

Answer to Problem 21.65AP

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

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

In the process from point $\text{A}$ to point $\text{C}$, the pressure does not changes.

The expression for the process from point $\text{A}$ to point $\text{C}$ is,

$\frac{V_{\text{A}}}{T_{\text{A}}}=\frac{V_{\text{C}}}{T_{\text{C}}}$

Here,

$V_{\text{C}}$ is the volume at point $\text{C}$

$T_{\text{C}}$ is the temperature at point $\text{C}$.

Substitute $5.00\text{L}$ for $V_{\text{A}}$, $300\text{K}$ for $T_{\text{A}}$ and $900\text{K}$ for $T_{\text{C}}$ in above equation.

$\begin{array}{c}\frac{5.00\text{L}}{300\text{K}}=\frac{V_{\text{C}}}{900\text{K}}\\ V_{\text{C}}=15.0\text{L}\end{array}$

Conclusion:

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

To determine

The experimental methods to carry out the process $A\to B$, $B\to C$ and $C\to A$.

Answer to Problem 21.65AP

The experimental method to carry out the process $A\to B$ is locking the piston, for process $B\to C$ is releasing the piston to expand at constant temperature, and for process $C\to A$ is releasing the piston to move freely.

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

In the process from point $\text{A}$ to point $\text{B}$:

The volume does not change. The temperature varies from $300\text{K}$ to $900\text{K}$ between these points. This process can be obtained by locking the piston so that the volume o gas does not changes and pressure increase from $1\text{atm}$ to $3\text{atm}$.

In the process from point $\text{B}$ to point $\text{C}$:

The temperature does not change. The pressure varies from $3\text{atm}$ to $1\text{atm}$ between these points. This process can be obtained by releasing the piston to expand at constant temperature.

In the process from point $\text{C}$ to point $\text{A}$:

The pressure does not change. The temperature varies from $900\text{K}$ to $300\text{K}$ between these points. This process can be obtained by releasing the piston to move freely so that after volume varies from $15\text{L}$ to $5\text{L}$.

Conclusion:

Therefore, the experimental method to carry out the process $A\to B$ is locking the piston, for process $B\to C$ is releasing the piston to expand at constant temperature, and for process $C\to A$ is releasing the piston to move freely.

To determine

The heat $Q$, the work done $W$ and change in internal energy $\Delta E_{\mathrm{int}}$ for each of the process.

Answer to Problem 21.65AP

The heat $Q$, the work done $W$ and change in internal energy $\Delta E_{\mathrm{int}}$ for process $A\to B$ are $1.52\text{kJ}$, 0, $1.52\text{kJ}$ respectively and for process $B\to C$ are $1.67\text{kJ}$, $-1.67\text{kJ}$,0 respectively and for process $C\to A$ are $-2.53\text{kJ}$, $1.01\text{kJ}$, $-1.52\text{kJ}$.

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

For the process from point $\text{A}$ to point $\text{B}$:

The volume of gas does not change.

The work done is,

$W_1=0$

The change in internal energy is equal to the heat.

The expression for the change in internal energy is,

$\begin{array}{c}\Delta E_{{\mathrm{int}}_1}=Q_1\\ =n{C}_{\text{v}}\left(T_{\text{B}}-T_{\text{A}}\right)\end{array}$

Here,

$n$ is the number of moles.

$C_{\text{v}}$ is the coefficient of heat transfer at constant volume.

Substitute $\frac{3}{2}R$ for $C_{\text{v}}$, $0.203$ for $n$, $900\text{K}$ for $T_{\text{B}}$ and $300\text{K}$ for $T_{\text{A}}$ in above equation.

$\begin{array}{c}\Delta E_{{\mathrm{int}}_1}=0.203\times \frac{3}{2}R\times \left(900\text{K}-300\text{K}\right)\\ =182.7R\end{array}$

Substitute $8.314\text{J}/\text{mol}\text{.K}$ for $R$ in above equation.

$\begin{array}{c}\Delta E_{{\mathrm{int}}_1}=182.7\times 8.314\text{J}/\text{mol}\text{.K}\\ =1518.96\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =1.52\text{kJ}\end{array}$

Thus, change in internal energy in process from point $\text{A}$ to point $\text{B}$ is $1.52\text{kJ}$.

For the process from point $\text{B}$ to point $\text{C}$:

The temperature does not change.

The change in internal energy is,

$\Delta E_{{\mathrm{int}}_2}=0$

The expression of the work done is,

$\begin{array}{c}W=-Q_2\\ =-nR{T}_{\text{B}}\ln \left(\frac{V_{\text{C}}}{V_{\text{B}}}\right)\end{array}$

Substitute $8.314\text{J}/\text{mol}\text{.K}$ for $R$, $0.203$ for $n$, $900\text{K}$ for $T_{\text{B}}$, $15\text{L}$ for $V_{\text{C}}$ and $5\text{L}$ for $V_{\text{B}}$ in above equation.

$\begin{array}{c}{W}_2=-0.203\times 8.314\text{J}/\text{mol}\text{.K}\times 900\text{K}\times \ln \left(\frac{15\text{L}}{5\text{L}}\right)\\ =-1668.76\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =-1.67\text{kJ}\end{array}$

Thus the change in internal energy in process from point $\text{B}$ to point $\text{C}$ is $-1.67\text{kJ}$.

For the process from point $\text{C}$ to point $\text{A}$:

The formula of work done is,

$W_3=P_{\text{A}}\left(V_{\text{C}}-V_{\text{A}}\right)$

Substitute $1\text{atm}$ for $P_{\text{A}}$, $15\text{L}$ for $V_{\text{C}}$ and $5\text{L}$ for $V_{\text{A}}$ in above equation.

$\begin{array}{c}{W}_3=1\text{atm}\left(\frac{101\times {10}^3\text{pa}}{1\text{atm}}\right)\times \left(15\text{L}-5\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\\ =1010\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =1.01\text{kJ}\end{array}$

Thus, the work done for the point $\text{C}$ to point $\text{A}$ is $1.01\text{kJ}$.

The formula for the change in kinetic energy is,

$\Delta E_{{\mathrm{int}}_3}=\frac{3}{2}nR\left(T_{\text{A}}-T_{\text{C}}\right)$

Substitute $8.314\text{J}/\text{mol}\text{.K}$ for $R$, $0.203$ for $n$, $900\text{K}$ for $T_C$ and $300\text{K}$ for $T_{\text{A}}$ in above equation.

$\begin{array}{c}\Delta E_{{\mathrm{int}}_3}=\frac{3}{2}\times 0.203\times 8.314\text{J}/\text{mol}\text{.K}\times \left(300\text{K}-900\text{K}\right)\\ =-1518.97\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =-1.52\text{kJ}\end{array}$

The heat obtain in this process is,

$\begin{array}{c}{Q}_3=\Delta E_{{\mathrm{int}}_3}-W_3\\ =-1.51\text{kJ}-1.01\text{kJ}\\ =-2.53\text{kJ}\end{array}$

Conclusion:

Therefore, the heat $Q$, the work done $W$ and change in internal energy $\Delta E_{\mathrm{int}}$ for process $A\to B$ are $1.52\text{kJ}$, 0, $1.52\text{kJ}$ respectively and for process $B\to C$ are $1.67\text{kJ}$, $-1.67\text{kJ}$,0 respectively and for process $C\to A$ are $-2.53\text{kJ}$, $1.01\text{kJ}$, $-1.52\text{kJ}$.

To determine

The heat $Q$, work done $W$ and $\Delta E_{\mathrm{int}}$ for the whole cycle $A\to B\to C\to A$.

Answer to Problem 21.65AP

For the whole cycle $A\to B\to C\to A$, the heat $Q$, work done $W$ and $\Delta E_{\mathrm{int}}$ are $0.656\text{kJ}$, $-0.656\text{kJ}$ and $0$ respectively.

Explanation of Solution

Given info: The volume of the monatomic ideal gas is $5.00\text{L}$ at atmospheric pressure and $300\text{K}$ temperature.

The expression for the heat in complete cycle is,

$Q=Q_1+Q_2+Q_3$

Substitute $1.51\text{kJ}$ for $Q_1$, $1.66\text{kJ}$ for $Q_2$ and $-2.52\text{kJ}$ for $Q_3$ in above equation.

$\begin{array}{c}Q=1.51\text{kJ}+1.66\text{kJ}-2.52\text{kJ}\\ =0.65\text{kJ}\end{array}$

Thus, the heat in cycle is $0.65\text{kJ}$.

The expression for the work done in complete cycle is,

$W=W_1+W_2+W_3$

Substitute $0$ for $W_1$, $-1.66\text{kJ}$ for $W_2$ and $1.01\text{kJ}$ for $W_3$ in above equation.

$\begin{array}{c}W=0-1.66\text{kJ}+1.01\text{kJ}\\ =-0.65\text{kJ}\end{array}$

Thus, the total work done is $-0.65\text{kJ}$.

As the process is cyclic, the change in internal energy will be zero.

$\Delta E_{\mathrm{int}}=0$

Conclusion:

Therefore, For the whole cycle $A\to B\to C\to A$, the heat $Q$, work done $W$ and $\Delta E_{\mathrm{int}}$ are $0.656\text{kJ}$, $-0.656\text{kJ}$ and $0$ respectively.