91Ó°ÊÓ

For the cyclic process$\underset{cyde}{∑}E=0$ and for ideal gas, the internal energy does not change when the temperature does not change. Also, in an adiabatic process, heat. By using these concepts, we can find the change in internal energy of the gas along the path CD.

For the cyclic process

$\underset{cyde}{∑}E=0$

In adiabatic process

$\mathrm{Δ}E=-W$

1. The path AB is isothermal at 300k
2. The path BC is adiabatic with work W = 5.0J
3. The path CD is at a constant pressure of 5 atm
4. The path DE is isothermal
5. The path EA is adiabatic with change in the internal energy of $\mathrm{Δ}{\text{E}}_{E→A}=8.0\text{J}$.

For an adiabatic process, first law of thermodynamics gives-

$\mathrm{Δ}E=-W………………………………..\left(18-28\right)$

Where $\mathrm{Δ}E$is the change in internal energy and w is work done

we know that for a cyclic process, the internal energy is conserved. This gives-

role="math" localid="1661861560615" $\mathrm{Δ}{E}_{A→B}+\mathrm{Δ}{E}_{B→C}+\mathrm{Δ}{E}_{C→D}+\mathrm{Δ}{E}_{D→E}+\mathrm{Δ}{E}_{E→A}=0$

Where $\mathrm{Δ}{E}_{A→B},\mathrm{Δ}{E}_{B→C},\mathrm{Δ}{E}_{C→D},\mathrm{Δ}{E}_{D→E}and\mathrm{Δ}{E}_{E→A}$are the change in the internal energies from $AtoB,BtoC,CtoD,DtoEandDtoA$ respectively.

Since ideal gas is involved, the internal energy does not change when the temperature does not change, so

$\mathrm{Δ}{E}_{A→B}=\mathrm{Δ}{E}_{D→E}=0$

Now, with$\mathrm{Δ}{E}_{E→A}=8.0\text{J}$, we have

$0+\mathrm{Δ}{E}_{B→C}+\mathrm{Δ}{E}_{C→D}+0+8.0\text{J}=0$

In the adiabatic process,

$\mathrm{Δ}E=-W……………………………………………………………………………………..\left(18-28\right)$

And$W=5.0J$.

Therefore, we get

$$\begin{gathered} -5.0\text{J}+\mathrm{Δ}{E}_{C→D}+0+8.0\text{J}=0 \\ \mathrm{Δ}{\text{E}}_{C→D}=-3.0\text{J} \end{gathered}$$

The change in internal energy of the gas in process C to D is$\mathrm{Δ}{\text{E}}_{C→D}=-3.0\text{J}$ .