We know that for an ideal gas

$$PV=nRT,$$

where $P$ is pressure, $V$ volume, $n$ amount of substance in moles, and $T$ the temperature of the gas.

We can easily derive that, for a (non-free) adiabatic expansion, the product $PV^γ=\text{const.} \implies TV^{γ-1}=\text{const.}$

From the first expression we see that we can vary $V$ and $P$ such that $T$ remains constant. But from $TV^{γ-1}=\text{const.} \implies T=\frac{\text{const.}}{V^{γ-1}}$, since $V$ has varied, the temperature should not remain constant. What am I missing?

Therefore and isothermic process cannot be an adiabatic one. Is this true?

## Best Answer

The above statement is wrong.

The ideal gas equation is as follows:

$$PV = nRT$$

If a process is isothermal, then $PV$ must remain constant.

If a process is adiabatic, it must satisfy $PV^{\gamma} = \text{const.}$

The above two conditions cannot be satisfied together.

Therfore, an isothermal process cannot be adiabatic process and vice versa.