Dry adiabatic lifting If an unsaturated air parcel ascends adiabatically in the atmosphere with an initial temperature, would the final temperature of the air parcel, when it reaches it's destination, be the same at the initial temperature since it does not exchange heat with it's surroundings?
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}