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An exercise in my book states the following:

n = 0.42 moles of an ideal biatomic gas are stored in an adiabatic cylinder with volume $V = 10^{-2} \text{ m}^3$, with an initial pressure equal to one atmosphere $p_\text{atm}$. A small solid object at a temperature $T_0 = 580 \text{ K}$ is placed very quickly inside the cylinder; its volume is negligible when compared to $V$. Once the temperature has reached its equilibrium at $T_1$ the object is removed very quickly, and one of the cylinder's bases is let go so that it can freely slide until the gas reaches a new equilibrium at a pressure $p_\text{atm}$ equal to the one applied externally, which has remained constant through the whole process, during which it does a work equal to $W = 705.4 \text{ J}$. Calculate the object's thermal capacity and the change in the gas' entropy.

In the solutions at the end of the book, it finds the initial temperature of the gas to be $T = 289.2 \text{ K}$ using the ideal gas law, and then it says that the expansion is adiabatic since there is no heat transfer, and that it can be assumed to be reversible.
The solution then states that the work done by the gas during the adiabatic expansion is $$ W = p_\text{atm} (V_2 - V) $$ I understand that the pressure outside the cylinder is constant, but since this is the work done by the gas shouldn't one consider the gas's pressure instead of the external one?\ I don't understand why work is calcualted that way, since pressure varies in an adiabatic expansion, and if that expansion is reversible it should hold that $$ p_f = \left(\frac{V_i}{V} \right)^{\gamma} p_i $$ The solutions then also claim that the work done is $$ W = -n c_v (T_2 - T_1) $$ Where $T_2$ is the final temperature after the expansion. This is what I expected the correct expression for the work to be, as it is the opposite of the change in the internal energy.
So are the solutions wrong? And if they aren't, why is the work equal to the one done in an isobaric expansion?

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  • $\begingroup$ Do you think that the gas satisfies the ideal gas law in an irreversible expansion against a constant external pressure? $\endgroup$
    – Chet Miller
    Commented Jul 7 at 15:12
  • $\begingroup$ The solutions say it is reversible, either way, even if it wasn't, It wouldn't follow the ideal gas law, but I also don't see why it would behave as if it was constant $\endgroup$
    – NICOLA TROMBINI
    Commented Jul 8 at 14:52
  • $\begingroup$ Suppose you have a massless, frictionless piston. Then the force balance on the piston is $F_{gas}-P_{ext}A=ma=0$ or $F_{gas}=P_{ext}A$, so work done by gas is $W_{gas}=P_{est}\Delta V$ $\endgroup$
    – Chet Miller
    Commented Jul 8 at 15:02
  • $\begingroup$ That makes sense, but why can the acceleration be null? Are you considering the average acceleration? Since I'd expect the acceleration to be positive in the initial phase and negative in the final phase when the piston slows down. Also, the reason this can't be applied with reversible transformations is that the external pressure would then equal the internal one, which is still given by Poisson's equations, correct? $\endgroup$
    – NICOLA TROMBINI
    Commented Jul 9 at 13:15
  • $\begingroup$ The acceleration is no null. The mass of the piston is zero and the acceleration of the piston is finite (so the kinetic energy of the piston is zero at all times). Even if the piston has finite mass, in the end (as you point out), its kinetic energy of the piston is zero; so, in this case, the energy balance on the piston again reduces to $W_{gas}=\P_{ext}\Delta V$. I don't understand the reference to Poisson. We have two different gases on either side of the piston, and Poisson's condition of isotropy only applies at static equilibrium. $\endgroup$
    – Chet Miller
    Commented Jul 9 at 13:31

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