An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in Figure 1.10(b). Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are “frozen out.” Also, assume that the only type of work done on the gas is quasistatic compression-expansion work.

1. For each of the four steps $\mathit{A}$ through $\mathit{D}$, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of ${\mathit{P}}_{\mathbf{1}}$, ${\mathit{P}}_{\mathbf{2}}$, ${\mathit{V}}_{\mathbf{1}}$, and ${\mathit{V}}_{\mathbf{2}}$. (Hint: Compute role="math" localid="1651641251162" $\mathbf{\Delta }\mathit{U}$before $\mathit{Q}$, using the ideal gas law and the equipartition theorem.)
   
2. Describe in words what is physically being done during each of the four steps; for example, during step A, heat is added to the gas (from an external flame or something) while the piston is held fixed.
   
3. Compute the net work done on the gas, the net heat added to the gas, and the net change in the energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.

Expression for work done when volume varies is

$W=-\underset{V_i}{\overset{V_f}{\int }}PdV$ …… (1)

Here, $P$is the pressure of the gas, $V_i\text{\hspace{0.17em}and\hspace{0.17em}}{V}_f$are the initial and final volume.

If a system contains $N$molecules, each with $f$the degree of freedom and there are no other temperature-dependent forms of energy then its total thermal energy is

$U_{thermal}=N.f.\frac{1}{2}kT$ …… (2)

Technically it is just the average total thermal energy. But if $N$is large then the fluctuation away from the average will be negligible.

The first law of thermodynamics can be expressed as

$Q=\Delta U-W$ …… (3)

Here,$Q$ is the amount of heat added,$\Delta U$ is the net change in thermal energy, and$W$ is work done on the gas.

The gas in this case is diatomic but the temperature is low enough that only translation $\left(f=3\right)$and $\left(f=2\right)$rotational degrees of freedom are excited so vibration modes are frozen out.

Work done for side A (since volume is constant along the side A, $dV=0$) is

$\boxed{W_A=0}$ …… (4)

For part B, volume is constant throughout so work done is

$W_B=\underset{V_1}{\overset{V_2}{\int }}{P}_{2}dV$

$W_B=-P_2\left(V_2-V_1\right)$ …… (5)

Here, pressure is constant on this site and $V_2>V_1$.

On the other hand, work done on side D where volume is constant is

$\boxed{W_D=0}$ …… (6)

For side C, the equation for pressure is

$\begin{array}{l}{W}_C=-P_1\left(V_1-V_2\right)=P_1\left(V_2-V_1\right)\\ \boxed{W_C=P_1\left(V_2-V_1\right)}\end{array}$ …… (7)

The thermal energy of the gas is

$U=\frac{5}{2}NkT=\frac{5}{2}PV$ …… (8)

Along the side A, because the volume is constant, thermal energy is

$\Delta U_A=\frac{5}{2}V\Delta P$

$\boxed{\Delta U_A=\frac{5}{2}{V}_1\left(P_2-P_1\right)}$ ……. (9)

Alongside B, since the pressure is constant, thermal energy is therefore

$\Delta U_B=\frac{5}{2}P\Delta V=\frac{5}{2}{P}_2\left(V_2-V_1\right)$

$\boxed{\Delta U_B=\frac{5}{2}{P}_2\left(V_2-V_1\right)}$ …… (10)

Alongside D, volume is constant so the thermal energy is therefore

$\Delta U_D=\frac{5}{2}V\Delta P=\frac{5}{2}{V}_1\left(P_1-P_2\right)$

$\boxed{\Delta U_D=-\frac{5}{2}{V}_2\left(P_2-P_1\right)}$ …… (11)

Alongside C, the pressure is constant so the thermal energy is therefore

$\Delta U_C=\frac{5}{2}P\Delta V=\frac{5}{2}{P}_1\left(V_1-V_2\right)$

$\boxed{\Delta U_C=-\frac{5}{2}{P}_1\left(V_2-V_1\right)}$ …… (12)

The amount of heat added can be obtained from the first law of thermodynamics from equation (3),

Substitute $0\text{\hspace{0.17em}J}$for $W_A$, $\frac{5}{2}{V}_1\left(P_2-P_1\right)$for $\Delta U_A$in equation (3) heat alongside A can be obtained

$Q_A=\frac{5}{2}{V}_1\left(P_2-P_1\right)-0$

$\boxed{Q_A=\frac{5}{2}{V}_1\left(P_2-P_1\right)}$ …… (13)

Substitute $-P_2\left(V_2-V_1\right)$for $W_B$, $\frac{5}{2}{P}_2\left(V_2-V_1\right)$for $\Delta U_B$in equation (3) heat alongside B is calculated as

$Q_B=\frac{5}{2}{P}_2\left(V_2-V_1\right)+P_2\left(V_2-V_1\right)$

$\boxed{Q_B=\frac{7}{2}{P}_2\left(V_2-V_1\right)}$ …… (14)

Substitute for $0\text{\hspace{0.17em}J}$$W_D$, $-\frac{5}{2}{V}_2\left(P_2-P_1\right)$for $\Delta U_B$in equation (3) heat alongside D is calculated

$\begin{array}{c}{Q}_D=-\frac{5}{2}{V}_2\left(P_2-P_1\right)+0\\ =-\frac{5}{2}{V}_2\left(P_2-P_1\right)\end{array}$

$\boxed{Q_D=-\frac{5}{2}{V}_2\left(P_2-P_1\right)}$ …… (14)

Substitute$P_1\left(V_2-V_1\right)$ for $W_C$,$-\frac{5}{2}{P}_1\left(V_2-V_1\right)$ for$\Delta U_C$ in equation (3) heat alongside D is calculated

$\begin{array}{c}{Q}_C=-\frac{5}{2}{P}_1\left(V_2-V_1\right)-P_1\left(V_2-V_1\right)\\ =-\frac{7}{2}{P}_1\left(V_2-V_1\right)\end{array}$

$\boxed{Q_C=-\frac{7}{2}{P}_1\left(V_2-V_1\right)}$ …… (15)

Hence, work done on sides A, B, C, and D are$0\text{\hspace{0.17em}J,\hspace{0.17em}}-P_2\left(V_2-V_1\right),P_1\left(V_2-V_1\right)\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}J}$ respectively.

Head added to the gas on side A is $\frac{5}{2}{V}_1\left(P_2-P_1\right)$, along B is $\frac{7}{2}{P}_2\left(V_2-V_1\right)$, along C is and$-\frac{7}{2}{P}_1\left(V_2-V_1\right)$ alongside D is $-\frac{5}{2}{V}_2\left(P_2-P_1\right)$.

Work done on sides A, B, C, and D is$0\text{\hspace{0.17em}J,\hspace{0.17em}}-P_2\left(V_2-V_1\right),P_1\left(V_2-V_1\right)\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}J}$ respectively.

Head added to the gas on side A is $\frac{5}{2}{V}_1\left(P_2-P_1\right)$, along B is $\frac{7}{2}{P}_2\left(V_2-V_1\right)$, along C is$-\frac{7}{2}{P}_1\left(V_2-V_1\right)$ and alongside D is $-\frac{5}{2}{V}_2\left(P_2-P_1\right)$.

Alongside A, no work is done but heat is added to the gas to increase the pressure. Alongside B, the gas expands but the heat must be added to achieve this. Similarly, alongside C, no work is done, and the gas gives off heat. Along with side D, work must be done on the gas to compress it and during this process, the gas gives off heat.

According to the first law of thermodynamics, the amount of heat is calculated by subtracting the total net work done from the total change in energy during the process where heat is defined as any spontaneous flow of energy from one object to another caused by a difference in temperature between the objects, whereas work can be defined as any other transfer of energy into the system.

The given figure is

Expression for work done when volume vary is

$W=-\underset{V_i}{\overset{V_f}{\int }}PdV$ …… (1)

Here, $P$is the pressure of the gas, $V_i\text{\hspace{0.17em}and\hspace{0.17em}}{V}_f$are the initial and final volume.

If a system contains $N$molecules, each with $f$degree of freedom and there are no other temperature-dependent forms of energy then its total thermal energy is

$U_{thermal}=N.f.\frac{1}{2}kT$ …… (2)

Technically it is just the average total thermal energy. But if $N$is large then the fluctuation away from the average will be negligible.

The first law of thermodynamics can be expressed as

$Q=\Delta U-W$ …… (3)

Here,$Q$ is the amount of heat added,$\Delta U$ is the net change in thermal energy, and$W$ is work done on the gas.

Net work done is calculated as

$W_{total}=W_A+W_B+W_C+W_D$ …… (4)

The calculation for network done, substitute $0\text{\hspace{0.17em}J}$for $W_A$, $-P_2\left(V_2-V_1\right)$for $W_B$, $P_1\left(V_2-V_1\right)$for $W_C$, and $0\text{\hspace{0.17em}J}$for $W_D$in equation (4)

$\begin{array}{l}{W}_{total}=0-P_2\left(V_2-V_1\right)+P_1\left(V_2-V_1\right)+0\\ \boxed{W_{total}=\left(V_2-V_1\right)\left(P_1-P_2\right)}\end{array}$

Net heat added to the gas is given by

$Q_{total}=Q_A+Q_B+Q_C+Q_D$ …… (5)

The calculation for net heat added to the system substitute$\frac{5}{2}{V}_1\left(P_2-P_1\right)$ for $Q_A$,$\frac{7}{2}{P}_2\left(V_2-V_1\right)$ for $Q_B$,$-\frac{7}{2}{P}_1\left(V_2-V_1\right)$ for $Q_C$, and$-\frac{5}{2}{V}_2\left(P_2-P_1\right)$ for$Q_D$ in equation (5)

$\begin{array}{l}{Q}_{total}=\frac{5}{2}{V}_1\left(P_2-P_1\right)+\frac{7}{2}{P}_2\left(V_2-V_1\right)-\frac{7}{2}{P}_1\left(V_2-V_1\right)-\frac{5}{2}{V}_2\left(P_2-P_1\right)\\ =\frac{5}{2}\left(V_1-V_2\right)\left(P_2-P_1\right)-\frac{5}{2}\left(P_1-P_1\right)\left(V_1-V_2\right)\\ =\left(V_2-V_1\right)\left(P_2-P_1\right)=-W_{total}\\ \boxed{Q_{total}=\left(V_2-V_1\right)\left(P_2-P_1\right)=-W_{total}}\end{array}$

Net energy change is calculated as

$\Delta U_{total}=\Delta U_A+\Delta U_B+\Delta U_C+\Delta U_D$ …… (6)

The calculation for the net change in energy, substitute$\frac{5}{2}{V}_1\left(P_2-P_1\right)$ for $\Delta U_A$,$\frac{5}{2}{P}_2\left(V_2-V_1\right)$ for $\Delta U_B$,$-\frac{5}{2}{V}_2\left(P_2-P_1\right)$ for$\Delta U_D$ and$-\frac{5}{2}{P}_1\left(V_2-V_1\right)$ for$\Delta U_C$ in equation (6)

$\begin{array}{c}\Delta U_{total}=\frac{5}{2}{V}_1\left(P_2-P_1\right)+\frac{5}{2}{P}_2\left(V_2-V_1\right)-\frac{5}{2}{V}_2\left(P_2-P_1\right)-\frac{5}{2}{P}_1\left(V_2-V_1\right)\\ =\frac{5}{2}\left(V_1-V_2\right)\left(P_2-P_1\right)+\frac{5}{2}\left(P_2-P_1\right)\left(V_2-V_1\right)\\ =\cancel{\frac{5}{2}\left(V_1-V_2\right)\left(P_2-P_1\right)}-\cancel{\frac{5}{2}\left(V_1-V_2\right)\left(P_2-P_1\right)}\\ =0\end{array}$

$\boxed{\Delta U_{total}=0}$.

The results are as expected. The net change in energy after going round all the four sides is zero since the gas comes back to its original position. Since the total heat is positive, it can be concluded that heat has been absorbed by the system. Work is done by the surrounding on the gas.

Hence, the network done on the gas is $W_{total}=\left(V_2-V_1\right)\left(P_1-P_2\right)$, net heat added to the gas is$Q_{total}=\left(V_2-V_1\right)\left(P_2-P_1\right)=-W_{total}$ and net energy change in this process is $\Delta U_{total}=0$.