AEstimate the ratio of multiplicities of water and water vap

A)Estimate the ratio of multiplicities of water and water vapor for this number of molecules knowing dS=dQ/T. Note: Consider a cubic cemtimeter and Qvaporization=40.66kj

What is the volume change would we need to account for the liquid and vapor multiplicity ratio?

Solution

a) In the thermodynamic limit

dS = dQ/T

We can find dQ (and then dS) by counting the number of water molecules in 1cc of water

dQ = NQ_V = 1cm³ (g/cm³)(mol/18g) Q_V = Q_V mol/18

dS = dQ /T = Q_V mol/18T = 4.066 × 10⁴J/mol x (mol/(18× 373K)) = 6.06J/K.

Comparing this macroscopic value for the entropy change with the microscopic form

S = S_G S_L = k_B ln (W_G/ W_L)

from which we see that the ratio of multiplicities is (W_G/ W_L) = exp (S/k_B)

= exp (6.06J/K /1.38 × 10²³J/K) = exp (4.39 × 10²³) = 10^{4.39×10^23/ ln 10} = 10^{1.9×10^23}

b) We have the relation between the numbers of states W and the volume occupied V

W_G/ W_L= (V_G/V_L)^N

where N is the number of molecules. For 1cc of water N = 1cm³ (g/cm³)(mol/18g) (6.24 × 10²³/mol)

= 3.3 × 10²²

Thus, the predicted change in volume is V_G/V_L = (W_G/W_L)^1/N = (10^{1.9×10^23})^{1/3.3×10^22} = 570000 = 6 × 10⁵ .