91Ó°ÊÓ

In this process, there is a transfer/exchange of energy, but there is no transfer/exchange of heat or mass from the specific system or surroundings. So, the value of heat in this process becomes zero.

The ideal gas equation for n mole gas is given by:

$$\begin{gathered} PV=nRT \\ V=\frac{nRT}{P} \end{gathered}$$

Here, P is the pressure, V is the volume, and T is the temperature.

For the adiabatic process, we have,

role="math" localid="1663829285439" $$\begin{gathered} P{V}^{\mathrm{γ}}=const. \\ P.{\left(\frac{nRT}{P}\right)}^{\mathrm{γ}}=const. \\ {P}^{1-\mathrm{γ}}{T}^{\mathrm{γ}}=const..............\left(i\right) \end{gathered}$$

If the pressure decrease in the adiabatic process, then,

$P_2<P_1$

Therefore, the adiabatic equations at different pressures and temperatures are constant, so compare them to find the relation.

$$\begin{gathered} P^{1-\mathrm{γ}}{T}^{\mathrm{γ}}=const. \\ {P}_1^{1-\mathrm{γ}}{T}^{\mathrm{γ}}=P_2^{1-\mathrm{γ}}{T}^{\mathrm{γ}} \\ {\left(\frac{T_1}{T_2}\right)}^{\mathrm{γ}}={\left(\frac{P_2}{P_1}\right)}^{1-\mathrm{γ}} \\ {\left(\frac{P_2}{P_1}\right)}^{\mathrm{γ}-1}={\left(\frac{T_1}{T_2}\right)}^{\mathrm{γ}} \\ \left(\frac{P_2}{P_1}\right)={\left(\frac{T_1}{T_2}\right)}^{\frac{\mathrm{γ}}{\mathrm{γ}-1}} \end{gathered}$$

Here, (1) and (2) are the two state points.

As we have, the pressure relation as:

$$\begin{gathered} P_2<P_1 \\ \frac{\mathrm{γ}}{\mathrm{γ}-1}>1..\left(ii\right) \end{gathered}$$

Therefore, the pressure canbe given as:

$$\begin{gathered} \frac{p_1}{p_2}>1 \\ {\left(\frac{T_1}{T_2}\right)}^{\frac{\mathrm{γ}}{\mathrm{γ}-1}}>1.....\left(iii\right) \end{gathered}$$

From the reference to the equation (ii) and equation (iii).

$T_1>T_2$

Therefore, the temperature decreases. The internal energy is related to the temperature change.

Hence, the internal energy of gas also decreases.