Isothermal vs. adiabatic compression of gas in terms of required energy

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?

• A better way to express the adiabatic work is $$W=-\frac{P_0V_0-PV}{\gamma-1}$$. – Satwik Pasani Nov 28 '13 at 14:06
• It should be noted that the expressions derived by the OP apply only to reversible processes, since s/he has assumed (without mentioning this explicitly) that $P_{ext} = P_{gas}$. Since this is designed for a course, this should be made explicit. – theorist Apr 18 at 23:36

To solve this, try to use what I call the "graphical apparatus". For an isothermal process: \begin{align} PV&=\text{constant}\\ P\mathrm{d}V&=-V\mathrm{d}P\\ \frac{\mathrm{d}P}{\mathrm{d}V}&=-\frac{P}{V}\\ \end{align} for adiabatic process: \begin{align} PV^\gamma&=\text{constant}\\ \frac{\mathrm{d}P}{\mathrm{d}V}&=-\gamma\frac PV \end{align} Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.

Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma>1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.

• But the work is given as $-\int PdV$ not as $\int PdV$ . – Koolman May 15 '18 at 16:51

$$W=nRT\ln{\frac{V_0}{V}}$$

You can make the comparison with the adiabatic case much more easily if you re-express the adiabatic work as: $$W=\frac{P_0V_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]=\frac{nRT_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]$$ But, for the isothermal case, $T=T_0$. So for the isothermal case, $$W=nRT_0\ln{\frac{V_0}{V}}$$ Now you can see how much more easily it would be to compare the two amounts of work.

• We can prove it wrong . See my comment on the above answer . – Koolman May 16 '18 at 17:38
• @Koolman PdV is the work done by the system on the surroundings. -PdV is the work done by the surroundings on the system. Do you disagree with this? – Chet Miller May 16 '18 at 18:19
• Yeah , according to chemistry conventions Work = $-\int PdV$ which is work done by gas (system ). – Koolman May 17 '18 at 2:22
• Hello Chester, I think there's some confusion here. Wrt the IUPAC 2007 document page 56: equation given is $\mathrm{d}U=\mathrm{d}Q+\mathrm{d}W$. This makes it clear that the work done by system is taken $\mathrm{d}W=-P\mathrm{d}V$ i.e. with the negative sign. The footnote also states: "$W>0$ indicates an increase in the energy of the system", which can only happen when we define $W=-\int P\mathrm{d}V$. This is also what Koolman was trying to convey in his comment to Satwik's answer. – Gaurang Tandon May 18 '18 at 8:22
• @Koolman See my comment to Gaurang Tandon detailing the error made in your cited reference. – Chet Miller May 18 '18 at 14:44

protected by Loong♦Dec 22 '17 at 7:16

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).