Dry adiabatic lifting
No. When the gas - let's assume it is an ideal gas - rises up it will expand because the pressure $p$ sinks the higher it comes. If it is allowed to expand quasi-statically under adiabatic conditions ($\delta Q = 0$) then it does work $W$ (pressure-volume work $W = - p \mathrm{d}V$ to be more precise) on its environment, and, hence, its internal energy $U$ is reduced, and its temperature $T$ changes because according to the first law of thermodynamics

\begin{align}
  \mathrm{d} U &amp;= \delta Q + \delta W \\
  &amp;\overset{\delta Q = 0}{=} \mathrm{d} W = - p \mathrm{d}V 
\end{align}

and also for an ideal gas you have

\begin{align}
  \mathrm{d} U &amp;= C_{V} \mathrm{d}T 
\end{align}

where $C_{V}$ is the heat capacity. Hence the temperature changes according to

\begin{align}
  \mathrm{d}T = - \frac{p}{C_{V}} \mathrm{d}V \ .
\end{align}