Let $ n = \dim X $. As $ X $ is rational, (by definition) $ X $ is birationally equivalent to $ \mathbb{P}^{n} $, and since the geometric genus $ p_{g}(X) (= P_{1}(X) = \dim H^{0} (X, \omega_{X} ) ) $ is a birational invariant, we have that $ p_{g} (X) = p_{g} (\mathbb{P}^{n}) $. But the canonical sheaf of $ \mathbb{P}^{n} $ is \[ \omega_{\mathbb{P}^{n}} \cong \mathscr{O}_{\mathbb{P}^{n}} (n1) , \]thus it has no (nontrivial) global sections. Hence \[ p_{g} (X) = p_{g} (\mathbb{P}^{n}) = \dim H^{0} ( \mathbb{P}^{n}, \omega_{\mathbb{P}^{n}} ) = 0. \]
