91Ó°ÊÓ

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 : $W=-{∫}_{V_i}^{V_f}PdV…….\left(1\right)$

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

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

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 :$Q=\mathrm{Δ}U-W…….\left(3\right)$

Here, Q is amount of heat added, $\mathrm{Δ}U$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 :$W_A=0……\left(4\right)$

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

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

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 : $W_D=0…….\left(6\right)$

For side C, the equation for pressure is :

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

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