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

The initial pressure is 1.5 atm. Since the rate of pressure drop is -dP/dt = kP, integration yields P(t) = P0 × e^-kt. Substituting P0 = 1.5 atm gives P(t) = 1.5 e^-kt atm.