Q1: Pure aniline is evaporating through a stagnant air film of 1 mm thickness at 300 K and a total pressure of 100 KPa. The vapor pressure of aniline at 300 K is 0.1 KPa. The total molar concentration under these conditions is 40.1 mole/m³ . The diffusivity of aniline in air is 0.74xl0^-5m²/s.The numerical value of mass transfer co-efficient is 7.4 x 10^-3. Its units are

ANS:C - mole/m².s.Pa

The mass transfer coefficient (𝑘k) is typically expressed in units of distance per time, such as meters per second (m/s) or centimeters per second (cm/s). Given that the diffusivity of aniline in air is provided in units of 𝑚2/𝑠m2/s and the numerical value of the mass transfer coefficient is 7.4×10−37.4×10−3, we can infer the units of the mass transfer coefficient. 𝑘=𝐷/𝐿k=D/L Where:

• 𝐷D is the diffusivity (𝑚2/𝑠m2/s)
• 𝐿L is the characteristic length (in this case, the thickness of the stagnant air film)

So, the units for 𝑘k will be the units of diffusivity divided by the units of the characteristic length. Given the diffusivity in 𝑚2/𝑠m2/s and the characteristic length (thickness of the stagnant air film) in meters, when you divide 𝑚2/𝑠m2/s by meters (𝑚m), you're left with 1/𝑠1/s, which corresponds to the units of reciprocal time. Therefore, the correct units for the mass transfer coefficient (𝑘k) are: 𝑚/𝑠m/s