An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the given figure.

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.

(a) For each of the four steps A through 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 $P_1,P_2,V_1,and{V}_2$. (Hint: Compute $\Delta U$before Q, using the ideal gas law and the equipartition theorem.)

(b) 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.

(c) 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.

An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the given figure.

We know formula for work done when volume varies is : $W=-{\int }_{V_i}^{V_f}PdV\dots \dots .\left(1\right)$

Here, P is pressure of the gas, $V_{i}and{V}_f$are 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 it total thermal energy is

$U_{\text{rhermal}}=N·f·\frac{1}{2}kT\dots \dots ..\left(2\right)$

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

First law of thermodynamics can be expressed as :$Q=\Delta U-W\dots \dots .\left(3\right)$

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

The gas in this case is diatomic but the temperature is low enough that only translation (f=3) and rotational (f=2) 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 :$W_A=0\dots \dots \left(4\right)$

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

$$\begin{gathered} W_B={\int }_{V_1}^{V_2}{P}_{2}dV \\ {W}_B=-P_2\left(V_2-V_1\right)\dots \dots ..\left(5\right) \end{gathered}$$

Here, pressure is constant at this side and $V_2>V_1.$

On the other hand, work done on side D where volume is constant is : $W_D=0\dots \dots .\left(6\right)$

For side C, the equation for pressure is :

$$\begin{gathered} W_C=-P_1\left(V_1-V_2\right)=P_1\left(V_2-V_1\right)\dots \dots \left(7\right) \\ {W}_C=P_1\left(V_2-V_1\right) \end{gathered}$$

Thermal energy of the gas is $U=\frac{5}{2}NkT=\frac{5}{2}PV$

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