Lets consider I have two vessels of the same volume $V$ each of which contains ideal gas at different pressures and temperatures $(p_1,T_1)$, $(p_2,T_2)$. I pump certain amount of the gas $ n_\Delta$ from one vessel to the other. What is the work required for the pumping (assuming adiabatic process, without friction or other losses).
Is there some elegant solution or simple equation using only pressure, temperatue and molar amount?
Not elegant solution:
We can use well known formula for adiabiatic process
$p V^\gamma = const = K$
We calculate volume fraction of each gas before and after from
so we calculate molar amounts in each vessel before pumping
$n_1 = p_1 V /( R T_1 )$
$n_2 = p_2 V /( R T_2 )$
and attribute volume fraction to each molar amount ($V \propto n$)
$V_1 = V(n_1-n_\Delta)/n_1 $ ... gas left in vessel 1
$V_2 = V$ ... gas already present in vessel 2
$\Delta V = V n_\Delta/n_1 $ ... gas pumped between vessels
Than we can calculate adiabatic compression/expansion for each volume independently $[V_1 \rightarrow V_1'; V_2 \rightarrow V_2'; V_\Delta \rightarrow V'_\Delta]$. To do so we have to determined final volume and pressure.
After the compression we have
$V_1' = V$
$V_2' + V'_\Delta = V $
$V'_\Delta = V n_\Delta /(n_2+n_\Delta) $
$V_2' = V n_2/(n_2+n_\Delta) $
Now we know volume of each fraction of gas, so we can use $p V^\gamma = const$ and
$W = (p'V' - pV)/(1-\gamma) $
... but this starts to be rather complicated ... so I wonder if there is more elegant way
NOTE : In addition I'm not sure about entropy when the temperature of pumped gas after compression is different than the gas originally in vessel ( => mixing gas of different temperatures)
EDIT : To make is a sketch of the pumping translated to compression by imaginary piston