For liquid na the 1 atm boiling temperature is 1150 k for a

Question

Problem -

a. Consider a one-component system consisting of spherical bubbles of vapor inside a liquid, which is commonly observed as a liquid begins to boil. This is similar to the case considered in class in which liquid droplets were suspended in a vapor, but the identity of the phases is reversed (here α = liquid, β = vapor). If the liquid is in equilibrium with the vapor in the bubbles, it can be shown that:

V^βdP^β = ΔH_v(dT/T)

where T is the boiling temperature. Beginning with this equation, derive the following equation showing the variation of boiling temperature with bubble size:

(1/T_o) - (1/T_r) = (R/ΔH_v)ln(1 + (2γ/rP^α))

Briefly explain any non-mathematical steps. Assume that the vapor can be treated as an ideal gas and state any other assumptions you make.

In this equation:

T_o = boiling temperature for a flat interface, K

T_r = boiling temperature for the liquid layer surrounding a bubble of radius r, K

ΔH_v = enthalpy of vaporization, J/mole

γ = surface energy of the liquid/bubble interface, J/m²

b. For liquid Na, the 1 atm boiling temperature is 1150 K for a flat liquid-vapor interface, ΔH_v = 101.3 kJ/mole and γ = 0.19 1/m². If the liquid pressure is 1 atm, calculate (i) the pressure (in atm) inside a Na bubble with a radius of 1 μm and (ii) the boiling temperature for the liquid layer surrounding a Na bubble with a radius of 1 μm.