Adiabatic expansion of ideal gas and dependence of temperature on volume

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?

• An adiabatic process is NOT isothermal process, instead an adiabatic and reversible process is isentropic. In fact an isothermal can never be adiabatic (reversible or irreversible), this is a consequence of Kelvin's formulation of 2nd law because it would imply that you could withdraw heat from a fixed temperature reservoir an perform work in a cycle. – hyportnex Jan 4 '16 at 12:59
• Thanks. Therefore the little argument of the question provides a proof that an adiabatic process can never be isothermal. Thanks. – Valentina Jan 4 '16 at 13:08
• @Valentina: the argument provides the proof of the statement for an ideal gas and for quasi-static processes. It does not prove the case of 'arbitrary substance' with quasi-static processes (where the statement is also true). For processes that are non-equilibrium, the statement is not true since expansion of an ideal gas into the vacuum is isothermal and adiabatic ($\Delta T=0$, $\Delta Q=0$). PS: I wonder why so many people ask this question? – ophelia Jan 4 '16 at 13:30
• For ideal gases, there are a handful of irreversible adiabatic processes (special cases) that are also isothermal: free expansion into a vacuum and Joule-Thompson throttling. – Chet Miller Jan 4 '16 at 14:06
• @ChesterMiller says right that an adiabatic expansion in vacuum is isothermal, but notice that this is an adiabitic process in the sense that there aren't heat exchanges, not that entropy is conserved: the entropy of the ideal gas is a known function of $V$ and $T$, so if $T$ is the same and $V$ changes also entropy has to change, because of the dependance on $V$. It happens because this process is irreversible. – Annibale Oct 2 '17 at 14:43

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.

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.

Simply yes an isothermal process cannot be adiabatic. A good way to think is an isothermal process is done by keeping the gas in thermal contact with an external bath of constant temperature, which means heat is being exhcanged which be definition means it is not adiabatic

$dQ=0$ implies $dU+pdV=0$ also.Now you want to change $p$ as well as $V$ such that $T$ is constant i.e $dU=0$.This will give $pdV=0$ which is contradictory to our assumption that we are changing $V$ giving $pdV\neq 0$.