Example 17: Calculate the electrode potential of given electrode $$\ce{Pt,Cl2 (\pu{1.5 bar}) | 2 Cl- (\pu{0.01 M})}\,; \quad E^\circ_\ce{Cl2/2Cl-} = \pu{1.36 V}$$
Solution: The reaction of electrode is $$\ce{\underset{\pu{1.5 bar}}{Cl2 (g)} + 2 e- -> \underset{\pu{0.01 M}}{2 Cl-}}$$ $$E = E^\circ - \frac{0.0591}{n}\log{\frac{[\ce{Cl-}]^2}{P_\ce{Cl2}}} = 1.36 - \frac{0.0591}{2}\log{\frac{(0.01)^2}{1.5}} = \pu{1.483 V}$$
In the solution, the author has directly used the pressure (which is in bars) in the Nernst equation without first converting it to its SI units. How can this be right?