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. The rate of evaporation of aniline is 2.97 x 10^-4. Its units are

ANS:D - kmole/m² . s

To find the units of the rate of evaporation of aniline, let's first understand the given data and equations involved. Given data:

• Thickness of stagnant air film (𝛿δ) = 1 mm = 0.001 m
• Temperature (𝑇T) = 300 K
• Total pressure (𝑃totalPtotal​) = 100 kPa = 100,000 Pa
• Vapor pressure of aniline (𝑃anilinePaniline​) = 0.1 kPa = 100 Pa
• Total molar concentration (𝐶totalCtotal​) = 40.1 mol/m³
• Diffusivity of aniline in air (𝐷anilineDaniline​) = 0.74×10−50.74×10−5 m²/s
• Rate of evaporation of aniline (𝑁N) = 2.97×10−42.97×10−4 mole/m²·s
• Mass transfer coefficient (𝑘k) = 7.4×10−37.4×10−3 m/s

The rate of evaporation (𝑁N) can be calculated using the equation for mass transfer through a stagnant film: 𝑁=𝑘⋅(𝑃total−𝑃aniline)⋅1𝛿N=k⋅(Ptotal​−Paniline​)⋅δ1​ Substituting the given values: 𝑁=(7.4×10−3 m/s)×(100,000 Pa−100 Pa)×10.001 mN=(7.4×10−3m/s)×(100,000Pa−100Pa)×0.001m1​ 𝑁=(7.4×10−3 m/s)×(99,900 Pa)×1000 mN=(7.4×10−3m/s)×(99,900Pa)×1000m 𝑁=7.4×10−3×99,900×1000 mol/m²\cdotpsN=7.4×10−3×99,900×1000mol/m²\cdotps 𝑁=7.4×10−1×99,900 mol/m²\cdotpsN=7.4×10−1×99,900mol/m²\cdotps 𝑁=7.4×104 mol/m²\cdotpsN=7.4×104mol/m²\cdotps So, the units of the rate of evaporation of aniline are mole/m²·s. Therefore, the correct option is: mole/m²·s