Question

e 9. [6 points] In a cell, chemical potential of phase 2 against phase 1 was measured at temperature T [K]. (ie, phase changed from phase 1 to phase 2) If this chemical potential is Δμ, what is the ratio of osmotic pressure between phase 2 and phase l (112 / П.) at T [K]? Assume that the activity of phase 1 and phase 2 can be expressed by its concentration C1 and C2, respectively. measured for phase l , ПГ osmotic pressure measured for phase 2) a) Δμ b) 1/Δμ c) RT In C2 C1 C2 d) -RTn e) eRT tfr f) eKT g) RT h) none of these

I know the answer is e but I don't understand how to get there.

Relevant equations may include:

$\Delta \mu = RTln(\frac{a_{1}}{a_{2}}) + ZFV$

$\Pi = cRT$

$ln(a_{a}) = \frac{-V_{a}*\Pi}{RT}$

Answer 1

We know pressure vary as following P proportionalto e^-E/K_B T because PV = NK_BT P proportionalto e^-E/RT PV = RT & K_B = R/N pi_1 = P_1 proportionalto e^-E_1/KT = C_1 e^-mu_1/RT because E_1 = mu_1 (chemical potential) & C_1 rightarrow const pi_2 = P_2 = C_1 e^-E_2/RT = C_1 e^-mu_2/RT pi_2/pi_1 = C_1 e^-mu_2/RT/C_2 e^-mu_1/RT = e^-(mu_2 - mu_1)/RT pi_2/pi_1 = e^+(mu_)1 - mu_2)/RT pi_2/pi_1 = e^delta u/RT because mu_1 > mu_2 & 0 mu = mu_1 - mu_2 because mu_1 > mu_2 because in initial phase No. of molecules will be larger than the final phase.