I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question:
If you have little energy available, would you rather use an isothermal or an adiabatic process to compress a gas?
My analysis is as follows:
The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=-\int_{V_0}^{V}PdV$$ For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above: $$W=-\int_{V_0}^{V}\frac{nRT}{V}dV=-nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V_0}{V}\right)$$ For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=-\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV=-\frac{P_0V_0^\gamma}{-\gamma+1}\left(V^{-\gamma+1}-V_0^{-\gamma+1}\right)=\frac{PV-P_0V_0}{\gamma-1}$$ $$W=-\frac{P_0V_0-PV}{\gamma-1}$$
This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy.
Now, finally, to my question:
Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed).
Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed?