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Yes indeed. The only consistent theory of matter we have today is quantum theory and quantum statistics. To establish the classical theory you have to borrow some ingredients from this more comprehensive theory. At least I know: (a) The natural measure of phase-space volumes in ##h^{d}=(2 \pi \hbar)^{d}## and (b) the indistinguishability of particles in terms of bosons and fermions (depending on spin).
The most simple way is to use quantum field theory ("2nd quantization") and the grand-canonical ensemble. As an example take spin-0 bosons. The one-particle states we take as given by the wave functions defined in a cubic volume (length ##L##) with periodic boundary conditions (in order to have properly defined momentum observables).
Then the quantum field is completely determined by the annihilation and creation operators for momentum eigenstates ##\hat{a}(\vec{p})## and ##\hat{a}^{\dagger}(\vec{p})## and the Hamilton operator is given by
$$\hat{H}=\sum_{\vec{p}} \frac{\vec{p}^2}{2m} \hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
With ##\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3##.
The annihilation and creation operators obey the commutation relations (bosonic fields)
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=\delta_{\vec{p},\vec{p}'}, \quad [\hat{a}(\vec{p}),\hat{a}(\vec{p}')]=0.$$
A convenient complete set of orthonomalized basis functions are the Fock states, i.e., the eigenstates of the occupation-number operators ##\hat{N}(\vec{p})=\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p})##. The eigenvalues are ##N(\vec{p}) \in \{0,1,2,\ldots\}=\mathbb{N}_0##.
To get the thermodynamics we need the grand-canonical partition sum
$$Z=\mathrm{Tr} \exp[-\beta (\hat{H}-\mu \hat{N})],$$
where
$$\hat{H}=\sum_{\vec{p}} E_{\vec{p}} \hat{N}(\vec{p}), \quad \hat{N}=\sum_{\vec{p}} \hat{N}(\vec{p}).$$
For the following it's more convenient to define the functional
$$Z[\alpha]=\mathrm{Tr} \exp[-\sum_{\vec{p}} \alpha(\vec{p}) \hat{N}(\vec{p})].$$
That's easy to calculate using the Fock basis (occupation-number basis)
$$Z(\alpha(\vec{p})=\prod_{\vec{p}} \sum_{N(\vec{p})=0}^{\infty} \exp[-\alpha(\vec{p}) N(\vec{p})] = \prod_{\vec{p}}\frac{1}{1-\exp(-\alpha(\vec{p})}.$$
The occupation number distribution is given by
$$f(\vec{p})=\langle \hat{N}(\vec{p}) \rangle=\frac{1}{Z} \mathrm{Tr} \hat{N}(\vec{p}) \exp[-\beta (\hat{H}=\mu)] .$$
This can be calculated from the functional
$$f(\vec{p})=-\left . \frac{\partial}{\partial \alpha(\vec{p})} \ln Z[\alpha] \right|_{\alpha(\vec{p})=\beta(E_{\vec{p}}-\mu)} = \frac{\exp[-\beta(E_{\vec{p}}-\mu)]}{1-\exp[-\beta(E_{\vec{p}}-\mu)]}=\frac{1}{\exp[\beta(E_{\vec{p}}-\mu)]-1}.$$
The partition sum itself is given by
$$\Omega(V,\beta,\mu)=\ln Z(V,\beta,\mu)=-\sum_{\vec{p}} \ln \{1-\exp[-\beta(E_{\vec{p}}-\mu)] \}.$$
The thermodynamic limit is not trivial since obviously we have the contraints ##\beta>0## and ##\mu<0##, and for too large ##\beta## and to large ##\mathcal{N}=\langle N \rangle## we cannot make ##L \rightarrow \infty## and keep ##n=\mathcal{N}/V## constant. The reason is that we need to treat the ground state ("zero mode" of the field) separately before doing the limit. The thorough investigation leads to the possibility of Bose-Einstein condensation for large ##n## and large ##\beta## (since ##\beta## turns out to be ##\beta=1/(k T)## that means low temperatures).
Restricting ourselves to non-degenerate states, i.e., high temperature and not too large ##n## we can naively make ##L \rightarrow \infty##. Then in any momentum-volume element ##\mathrm{d}^3 p## we have ##\frac{V}{(2 \pi \hbar)^3} \mathrm{d}^3 p## single-particle states and thus we can substitute the sum by an integral
$$\Omega=-\frac{V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 p \ln\{1-\exp[-\beta(E_{\vec{p}}-\mu)]\}.$$
The integral is non-trivial, but the classical limit is simple. That's given for small occupation numbers, i.e., for ##\exp[-\beta(E_{\vec{p}}-\mu]\ll 1##. Then we can set ##\ln(1-\exp(...))=-\exp(...)##
and
$$\Omega=\frac{V}{(2 \pi \hbar)^3} \int_{mathbb{R}^3} \mathrm{d}^3 p \exp[-\beta(E_{\vec{p}}-\mu)].$$
With ##E_{\vec{p}}=\vec{p}^2/(2m)## we can evaluate the Gaussian integral, leading to
$$\Omega=\frac{V}{(2 \pi \hbar)^3} \left (\frac{2 \pi m}{\beta} \right)^{3/2} \exp(\beta \mu).$$
Now the meaning of the constants become clear by evaluating the internal energy and the average particle number
$$\mathcal{N}=\langle N \rangle=\frac{1}{\beta} \partial_{\mu} \Omega=\Omega.$$
Further we have
$$U=\langle E \rangle=-\partial_{\beta} \Omega+ \mu \mathcal{N}=\frac{3}{2 \beta} \mathcal{N},$$
from which
$$\beta=1/(k T).$$
To get the relation to the more usual thermodynamic potentials we calculate the entropy. The statistical operator is
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H} + \beta \mu \hat{N})$$
and thus the entropy
$$S=-k \mathrm{Tr} \ln \hat{\rho}=-k (\Omega -\beta U + \beta \mu \mathcal{N})=-k \Omega+\frac{U-\mu}{T}.$$
To get the usual potentials we note that with
$$\Phi=\Phi(V,T,\mu)=-k T \Omega$$
one gets after some algebra
$$\mathrm{d} \Phi=\mathrm{d} V \partial_V \Phi - S \mathrm{d} T - \mathcal{N} \mathrm{d} \mu.$$
On the other hand from the above expression for the entropy we find
$$\Phi=U-S T - N \mu.$$
From this it follows
$$\mathrm{d} U = \mathrm{d} V \partial_V \Phi + T \mathrm{d} S+\mu \mathrm{d} \mathcal{N}$$
which gives
$$P=-\left (\frac{\partial \Phi}{\partial V} \right)_{T,\mu}.$$
The most simple way is to use quantum field theory ("2nd quantization") and the grand-canonical ensemble. As an example take spin-0 bosons. The one-particle states we take as given by the wave functions defined in a cubic volume (length ##L##) with periodic boundary conditions (in order to have properly defined momentum observables).
Then the quantum field is completely determined by the annihilation and creation operators for momentum eigenstates ##\hat{a}(\vec{p})## and ##\hat{a}^{\dagger}(\vec{p})## and the Hamilton operator is given by
$$\hat{H}=\sum_{\vec{p}} \frac{\vec{p}^2}{2m} \hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
With ##\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3##.
The annihilation and creation operators obey the commutation relations (bosonic fields)
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=\delta_{\vec{p},\vec{p}'}, \quad [\hat{a}(\vec{p}),\hat{a}(\vec{p}')]=0.$$
A convenient complete set of orthonomalized basis functions are the Fock states, i.e., the eigenstates of the occupation-number operators ##\hat{N}(\vec{p})=\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p})##. The eigenvalues are ##N(\vec{p}) \in \{0,1,2,\ldots\}=\mathbb{N}_0##.
To get the thermodynamics we need the grand-canonical partition sum
$$Z=\mathrm{Tr} \exp[-\beta (\hat{H}-\mu \hat{N})],$$
where
$$\hat{H}=\sum_{\vec{p}} E_{\vec{p}} \hat{N}(\vec{p}), \quad \hat{N}=\sum_{\vec{p}} \hat{N}(\vec{p}).$$
For the following it's more convenient to define the functional
$$Z[\alpha]=\mathrm{Tr} \exp[-\sum_{\vec{p}} \alpha(\vec{p}) \hat{N}(\vec{p})].$$
That's easy to calculate using the Fock basis (occupation-number basis)
$$Z(\alpha(\vec{p})=\prod_{\vec{p}} \sum_{N(\vec{p})=0}^{\infty} \exp[-\alpha(\vec{p}) N(\vec{p})] = \prod_{\vec{p}}\frac{1}{1-\exp(-\alpha(\vec{p})}.$$
The occupation number distribution is given by
$$f(\vec{p})=\langle \hat{N}(\vec{p}) \rangle=\frac{1}{Z} \mathrm{Tr} \hat{N}(\vec{p}) \exp[-\beta (\hat{H}=\mu)] .$$
This can be calculated from the functional
$$f(\vec{p})=-\left . \frac{\partial}{\partial \alpha(\vec{p})} \ln Z[\alpha] \right|_{\alpha(\vec{p})=\beta(E_{\vec{p}}-\mu)} = \frac{\exp[-\beta(E_{\vec{p}}-\mu)]}{1-\exp[-\beta(E_{\vec{p}}-\mu)]}=\frac{1}{\exp[\beta(E_{\vec{p}}-\mu)]-1}.$$
The partition sum itself is given by
$$\Omega(V,\beta,\mu)=\ln Z(V,\beta,\mu)=-\sum_{\vec{p}} \ln \{1-\exp[-\beta(E_{\vec{p}}-\mu)] \}.$$
The thermodynamic limit is not trivial since obviously we have the contraints ##\beta>0## and ##\mu<0##, and for too large ##\beta## and to large ##\mathcal{N}=\langle N \rangle## we cannot make ##L \rightarrow \infty## and keep ##n=\mathcal{N}/V## constant. The reason is that we need to treat the ground state ("zero mode" of the field) separately before doing the limit. The thorough investigation leads to the possibility of Bose-Einstein condensation for large ##n## and large ##\beta## (since ##\beta## turns out to be ##\beta=1/(k T)## that means low temperatures).
Restricting ourselves to non-degenerate states, i.e., high temperature and not too large ##n## we can naively make ##L \rightarrow \infty##. Then in any momentum-volume element ##\mathrm{d}^3 p## we have ##\frac{V}{(2 \pi \hbar)^3} \mathrm{d}^3 p## single-particle states and thus we can substitute the sum by an integral
$$\Omega=-\frac{V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 p \ln\{1-\exp[-\beta(E_{\vec{p}}-\mu)]\}.$$
The integral is non-trivial, but the classical limit is simple. That's given for small occupation numbers, i.e., for ##\exp[-\beta(E_{\vec{p}}-\mu]\ll 1##. Then we can set ##\ln(1-\exp(...))=-\exp(...)##
and
$$\Omega=\frac{V}{(2 \pi \hbar)^3} \int_{mathbb{R}^3} \mathrm{d}^3 p \exp[-\beta(E_{\vec{p}}-\mu)].$$
With ##E_{\vec{p}}=\vec{p}^2/(2m)## we can evaluate the Gaussian integral, leading to
$$\Omega=\frac{V}{(2 \pi \hbar)^3} \left (\frac{2 \pi m}{\beta} \right)^{3/2} \exp(\beta \mu).$$
Now the meaning of the constants become clear by evaluating the internal energy and the average particle number
$$\mathcal{N}=\langle N \rangle=\frac{1}{\beta} \partial_{\mu} \Omega=\Omega.$$
Further we have
$$U=\langle E \rangle=-\partial_{\beta} \Omega+ \mu \mathcal{N}=\frac{3}{2 \beta} \mathcal{N},$$
from which
$$\beta=1/(k T).$$
To get the relation to the more usual thermodynamic potentials we calculate the entropy. The statistical operator is
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H} + \beta \mu \hat{N})$$
and thus the entropy
$$S=-k \mathrm{Tr} \ln \hat{\rho}=-k (\Omega -\beta U + \beta \mu \mathcal{N})=-k \Omega+\frac{U-\mu}{T}.$$
To get the usual potentials we note that with
$$\Phi=\Phi(V,T,\mu)=-k T \Omega$$
one gets after some algebra
$$\mathrm{d} \Phi=\mathrm{d} V \partial_V \Phi - S \mathrm{d} T - \mathcal{N} \mathrm{d} \mu.$$
On the other hand from the above expression for the entropy we find
$$\Phi=U-S T - N \mu.$$
From this it follows
$$\mathrm{d} U = \mathrm{d} V \partial_V \Phi + T \mathrm{d} S+\mu \mathrm{d} \mathcal{N}$$
which gives
$$P=-\left (\frac{\partial \Phi}{\partial V} \right)_{T,\mu}.$$