# Ambient air compression question

Think of a long hollow cylinder (filled with ambient air)(say 80 degrees F) (at sea level) with a bottom and no top capable of fluid isolation X number of feet long. If I drop a piston into the cylinder how much heat and psi can I generate by the time the piston reaches equilibrium? The piston can be any weight to be able to reach maximum heat and temperature. The piston falls by gravity until it meets an equilibrium of force pushing the piston upwards equal to the weight of the piston forcing the piston downwards.

What is the maximum heat and pressure at this point? How far did I have to go down the cylinder and what was the final weight of the piston which I think will equal the psi.

When you drop a piston into a cylinder and the piston falls under its weight , the transformation can be fairly approximated as being adiabatic (no heat change with exterior). This is because the transformation happens suddenly and the time is not long enough to allow for heat exchange with surroundings. Also  the transformation IS NOT reversible which means the change in entropy (total disorder) will be greater than zero if you restore the cylinder+piston system to its initial state.

For the pressures things are simple

Initial pressure = P1 = atmospheric pressure

Final pressure= P2 = m*g/S + P1     (m is piston mass, S is piston area)

For heat
There are two terms in the heat produced Q1+Q2. First one corresponds to a reversible adiabatic transformation Q1= Q(reversible) and the second one Q2= Q(irreversible) corresponds to the nonzero charge in entropy (which gives the irreversibility of the transformation).

Q1 is simple to compute:

$P1*V1^\gamma = P2*V2^\gamma$     is the eq. for adiabatic process
$P1*V1 =N*R*T1$         , N is the number if gas moles
$P(V) = P1*(V1/V)^\gamma$      for air (bi atomic gas)
$\gamma= Cp/Cv = [(7/2)R]/[(5/2)R] =7/5$

$Q1 = \Delta(U) + W$       Heat = change in internal energy + Work done
$Q1 = N*Cv*(T2-T1) + \int_{V1}^{V2} [P(V)*dV] = NCv(T2-T1) +P1*V1*ln(V2/V1) =$

$=NCv(T2-T1) + P1*V1*ln(P1/P2)^{1/\gamma}$

To compute this further you need to know T2 qnd T1.

$T1 = P1*V1/(N*R)$

$T2 = P2*V2/(N*R)$

and $V2 = (P1/P2)^{1/\gamma}*V1$

$T2 = P2*(P1/P2)^{1/\gamma}*V1/(N*R)$

$Q1 = N*Cv*V1/(NR) [P1- (P2*P1/P2)^{1/\gamma}] + P1*V1*ln(P1/P2)^{1/\gamma}$

Now the heat Q2 corresponding to the irreversible process depends on the total variation of entropy $\Delta(S)$  after restoring the system to its initial state (or in other words how much disorder was created in the gas by dropping the piston down and restoring it back).

$Q2 = T1*\Delta(S)$

Q2 can not be computed further and the value of Q2 may change depending on the value of heat generated by piston friction, external heat losses (non perfect adiabatic process), etc.You need to specify exactly the amount of entropy (or disorder) created.

As a final response the heat created $Qtot >= Q1$ which was computed above.

You also can answer how far away the piston has come into the cylinder. If X1 is the initial length and X2 the final length of cylinder you have

$V2 = (P1/P2)^{1/\gamma}*V1$     (see above)
and

$X1/X2 = V1/V2 = (P2/P1)^{1/\gamma} = [(mg/S +P1)/P1]^{1/\gamma} =$

$= [1 +mg/(P1*S)]^{5/7}$