Reversible isothermal expansion and Reversible adiabatic expansion [closed]

For a given pressure drop, which of the two- isothermal reversible expansion and adiabatic reversible expansion - will produce a lesser volume increase?

closed as unclear what you're asking by Klaus-Dieter Warzecha, ringo, bon, Wildcat, JanSep 20 '16 at 9:49

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• It has been a long time since I took pchem but this doesn't make any sense to me. There seems to be too much left unspecified about the two systems. To me this is like saying I have two variables $x$, and $y$. Which is smaller? – MaxW Sep 19 '16 at 19:48
• But it is saying that the pressure drop is given, so cant we derive them using the thermodynamic equations? – Shivay Sep 19 '16 at 19:52
• I was just thinking about PV=nRT. For an isothermal expansion the whole expression nRT is a constant so fine. But for an adiabatic expansion T can can as well as V. – MaxW Sep 19 '16 at 20:04
• In the reversible adiabatic expansion, the temperature will decrease, while, in the reversible isothermal expansion, the temperature will stay constant. What does that tell you? – Chet Miller Sep 19 '16 at 20:09
• We need additional information. 1. Can we assume the gases are ideal? 2. Can we assume the expansions start and stop at the same pressures? – Curt F. Sep 19 '16 at 23:58

Let's assume that both expansions are for ideal gases.

Isothermal case

$$PV = \mathrm{constant} = c_{it}$$

$$P_1 V_1 = P_2 V_2$$

$$\frac{V_2}{V_1} = \frac{P_1}{P_2}$$

$$PV^\gamma = \mathrm{constant} = c_{it} = PV^\frac{f}{f+2}$$

Here, $\gamma$ is the ratio of heat capacities, and $f$ is the number of translational degrees of freedom of the molecule (3 for a mono-atomic gas, 5 for a diatomic gas, etc.). The expression involving $f$ makes clear that $\gamma$ must be $\gt 1$.

$$P_1 (V_1)^\gamma = P_2 (V_2)^\gamma$$

$$\left ( \frac{V_2}{V_1}\right)^\gamma = \frac{P_1}{P_2}$$

Comparing the two cases

For the same pressure drop, $\frac{P_1}{P_2}$ will be the same in each case. Say the pressure changes by two-fold, i.e. $\frac{P_1}{P_2}=2$. Then in the isothermal case, the volume will increase by two-fold. However in the adiabatic case, the volume will increase by a factor of $2^\gamma$. Since $\gamma \gt 1$, the volume increase will be more than two-fold.

Thus, the adiabatic case will have the greater volume increase, meaning the isothermal case will have the lesser volume increase.

The reason why

During adiabatic expansion, no energy is removed or supplied to the gas. However in isothermal expansion, that isn't true. To keep things isothermal during an expansion (an otherwise cooling process), heat would have to be continually added. Hotter gases take up more volume than cooler gases (if they are ideal). Thus the process of adding heat counteracts some of the volume change that could have happened if things were adiabatic. Thus the isothermal case results in a smaller volume change.