What will be the expression for pressures of gas above mercury in the closed arm at any time, t? — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Initial pressure $P = 1 \text{ atm} + 38 \text{ cm Hg} = 1.5 \text{ atm}$. From $-\frac{dP}{dt} = KP$, integration yields $P(t) = P_{initial} e^{-Kt} = 1.5 e^{-Kt}$. Therefore, correct answer is B.