An ideal gas, molar heat capacity $C_V=\frac{5}{2}R,$ is expanded adiabatically against a constant pressure of $1\ \mathrm{atm}$ until it doubles in volume. Given initial temperature of $25\ \mathrm{^\circ C}$ and initial pressure of $5\ \mathrm{atm}$, calculate final $T$ (etc.).*

The author uses the First Law to calculate $T_2$ as $274\ \mathrm K$.

My question is, why can't we use the relation $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$?

It appears we cannot, since $\gamma =C_p/C_V$ and so, since $V_2=2V_1$, $T_2=274\ \mathrm K$, $T_1=298\ \mathrm K$,

$$\log_2\frac{T_1}{T_2}+1 =\gamma = \frac{C_p}{5/2}\rightarrow C_p\approx 2.8, $$

so assuming $C_p - C_V = R$, giving $C_p = \frac{7}{2}R$, does not seem to work.

Can someone explain why the highlighted equation doesn't apply?

*This is adapted from an example in Castellan, Physical Chemistry.


1 Answer 1


The equations you are proposing to use are valid strictly for an adiabatic reversible expansion. In the present adiabatic expansion, the external force per unit area is dropped suddenly from 5 atm to 1 atm, and held at that value until the volume doubles. This takes place rapidly and spontaneously, and is not reversible, so the equations for an adiabatic reversible expansion do not hold. For an irreversible expansion, the ideal gas law (or other equation of state) cannot be used to describe the force per unit area acting on the piston face, because the equation of state only applies to a system in thermodynamic equilibrium (or, since a reversible expansion consists of a continuous sequence of thermodynamic equilibrium states, to a reversible expansion). In an irreversible expansion, the force at the piston face (where work is occurring) is determined not only by the amount of volume change, but by the rate at which the volume is changing. This is because of viscous stresses that are important in rapid expansions and compressions. In addition, the mass (inertia) of the gas comes into play, causing the pressure and temperature to be spatially non-uniform within the gas during its deformation. So there is no single uniform value of the pressure or the temperature that can be assigned to the gas.

  • $\begingroup$ I wonder if the definition of "adiabatic" used to include reversibility? Fermi's text, originally published in 1937, says on page 25: A transformation of a thermodynamical system is said to be adiabatic if it is reversible and if the system is thermally insulated..." $\endgroup$
    – daniel
    Commented Dec 3, 2017 at 5:51
  • $\begingroup$ Interesting. This surprises me. Incidentally, Fermi's text is excellent in specifying that, in using Clausius' Inequality, the temperature to employ in dQ/T is required to be that at the interface between the system and its surroundings (typically a reservoir) where the heat transfer is occurring. $\endgroup$ Commented Dec 3, 2017 at 13:34
  • $\begingroup$ It's a great text and I assume he meant to restrict reversibility to the example at hand (which uses the boxed relations above, so it clearly required reversibility). $\endgroup$
    – daniel
    Commented Dec 3, 2017 at 15:18

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