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For the pressure-density Hugoniot relation for a perfect gas, the change in specific entropy is $[s]/c_v=\log P-\gamma \log D(P)$, with $c_v$ the specific heat capacity at constant volume. Its derivative isIt is positive for a compressive normal shock wave with $P>1$. For a weak shock with $P=1+\delta$, integrating the leading term givesThus the entropy production is cubic in the small pressure jump, even though the mass density and temperature changes already appear at first order.

For $N$ equal weak shocks with individual pressure ratio $1+\delta$ and total ratio $P_s=(1+\delta {)}^N$, entropy production in a perfect-gas shock givesAt fixed $P_s>1$, this tends to zero as the compression is divided into increasingly many smaller weak shocks. A single finite normal shock wave instead gives the positive value $\log P_s-\gamma \log D(P_s)$. Gradual compression can consequently approach reversible isentropic flow while an abrupt compression generates finite entropy.

Work in the shock frame. The undisturbed upstream gas enters the stationary discontinuity at speedwhile the downstream gas, which moves with the piston in the laboratory frame, leaves atThe Rankine-Hugoniot conditions for a perfect gas areandEliminating the velocities between these equations gives the perfect-gas density jump

The mass and momentum equations also implysoSubstitution of the density ratio into mass conservation, followed by this momentum relation, givesUsing the adiabatic sound speed $c_1^2=\gamma p_1/{\rho }_1$ therefore yieldsFinally, eliminating $U$ from $V=\beta p_1/({\rho }_{1}U)$ gives the piston-driven normal shock relationFor a compressive shock $\beta >0$, the first boxed equation has $U>c_1$: the shock propagates supersonically into the undisturbed gas, approaching the sound speed in the weak shock limit $\beta \to 0$.

Put $S(P)=[s]/c_v$. Differentiating the expression in part (c) givesFor $P=1+\delta$, the denominator is $4{\gamma }^2+O(\delta )$, soIntegrating from the unshocked state, where $S(1)=0$, proves the cubic entropy production in a perfect-gas shock:The first-order and second-order terms vanish: a weak shock agrees with reversible compression through second order.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

The Taylor series of the shock density ratio at $\beta =0$ isFor a reversible adiabatic process, $p/p_0=(\rho /{\rho }_0{)}^{\gamma }$, so the density ratio predicted by the ideal-gas adiabat isEquations (1) and (2) agree through order ${\beta }^2$. Thus a weak shock satisfies the adiabatic pressure-density relation with an error of order $O({\beta }^3)$, which in particular establishes the requested accuracy through $O({\beta }^2)$.