A sphere of naphthalene of 6cm diameter is sublimated in stagnant air, Derive the equation.AND calculate the half-life.time of this sphere (the time it takes for the sphere to have a diameter of 3cm ). Additional information: Air is at 113F and 1atm. The molecular weight and density of naphthalene are 128g/mol and 71.1 lb/ft3, respectively. The vapor pressure of naphthalene at these conditions is 0.8mmHg. Diffusion of naphthalene in air at 45 " C is 6.92106m2/s. You will derive the equation starting from the two equations derived in class: dr=NAMWdt(1) NAT2=r0DABCln(1A1)(2) You have to replace (2) in (1) and solve the integral with limits t=0 to t=t1/2 and r=r0 to r=r0/2 This is the final solution: t31=(7ro2)(24DABCMWln(1yAiW1)) BE SURE THAT YOU UNDERSTAND HOW EQUATIONS (1) and (2) WERE DERIVED