Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid?
Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.

If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.