Question 2.11.3: An ideal gas is such that its internal energy is given by U ...

Since the process is reversible and adiabatic, there is no entropy variation. We can use the volume V as the unique state variable. Thus, the internal energy reads U(V) = c p (V) V.
The derivative of the internal energy with respect to the volume is given by,

\frac{dU}{dV}=c\frac{dp}{dV}V+cp=-p.

which can be recast as,

\frac{dp}{dp}+γ\frac{dV}{dV}=0.

where γ = (c + 1) /c. The integration of this expression from an initial state (p_i, V_i) to a final state (p_f, V_f) is written as,

\int_{p_{i}}^{p_{f}}{\frac{dp}{p} } +\gamma \int_{V_{i}}^{V_f}{\frac{dV}{V} } =0.

which yields,

thus
\ln \Biggl(\frac{p_fV^{\gamma }_{f} }{p_iV^{\gamma }_{i} } \Biggr)=0.

Therefore, the initial and final variables are related by,

p_iV^{\gamma }_{i}=p_fV^{\gamma }_{f}.

which yields the identity,

pV^γ=const.

The gas constant γ will be formally introduced in § 5.7.