What will be the increase in level of mercury in the closed arm at any time, t? — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Rise $x$ in closed arm causes fall $x/2$ in open arm. New diff $= 38 - 1.5x$. $P(t) = 114 e^{-Kt} = 114 - 1.5x \implies 1.5x = 114(1-e^{-Kt}) \implies x = 76(1-e^{-Kt})$. Therefore, correct answer is B.