= Solution
The one-particle <partition function> is
$$
Z_1=\int_0^\infty KV\varepsilon^{1/2}e^{-\beta\varepsilon}d\varepsilon
=\frac{KV\sqrt\pi}{2}\beta^{-3/2}.
$$
The classical indistinguishable noninteracting gas has $Z_N=Z_1^N/N!$; for labeled particles omit $N!$, which changes none of the requested <derivatives>. With $\beta=(k_BT)^{-1}$,
$$
\boxed{E=-\partial_\beta\log Z_N=\frac32Nk_BT,\qquad
C_V=\frac32Nk_B.}
$$
The allowed equation of state gives $PV=Nk_BT$, so the <enthalpy> is $E+PV=5Nk_BT/2$ and $\boxed{C_P=5Nk_B/2}$, at fixed particle number.
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