The rate of evaporation of water at steady state in a narrow metal tube is 1.21 x 10^-10 kg mol/(m2-s).
The rate of evaporation of water at steady state in a narrow metal tube can be calculated using Fick's first law of diffusion:
J = -D(∂c/∂z)
where:
J is the flux of water vapor (kg mol/(m2-s))
D is the diffusivity of water vapor in air (m2/s)
c is the concentration of water vapor in air (kg mol/m3)
z is the distance along the diffusion path (m)
The partial pressure of water vapor at the water surface is equal to the vapor pressure of water at 293 K, which is 0.0318 atm. The partial pressure of water vapor at the top of the tube is assumed to be zero. Therefore, the concentration gradient of water vapor is:
∂c/∂z = (0.0318 atm)/(0.1524 m) * (1 mol/L)/(1 atm) * (22.4 L/mol) = 0.00043 kg mol/m4
Substituting the values of D and ∂c/∂z into Fick's law, we get:
J = -(0.30 x 10^-4 m2/s) * (0.00043 kg mol/m4) = -1.29 x 10^-8 kg mol/(m2-s)
The negative sign indicates that the flux is in the direction of decreasing concentration, which is from the water surface to the top of the tube. The rate of evaporation is equal to the absolute value of the flux:
|J| = 1.29 x 10^-8 kg mol/(m2-s)
Therefore, the rate of evaporation of water at steady state in the narrow metal tube is 1.29 x 10^-8 kg mol/(m2-s).