- Mass Transfer - Section 1
- Mass Transfer - Section 2
- Mass Transfer - Section 3
- Mass Transfer - Section 4
- Mass Transfer - Section 5
- Mass Transfer - Section 6
- Mass Transfer - Section 7
- Mass Transfer - Section 8
- Mass Transfer - Section 9
- Mass Transfer - Section 10
- Mass Transfer - Section 11
- Mass Transfer - Section 12
- Mass Transfer - Section 13
- Mass Transfer - Section 14


Mass Transfer - Engineering
Q1: The value of NA/(NA + NB), for steady state equimolal counter diffusion of two gases 'A' and 'B' isA 1
B ∞
C 0.5
D 2
ANS:B - ∞ In steady-state equimolal counter-diffusion, the molar fluxes of the two gases, 𝑁𝐴NA and 𝑁𝐵NB, are equal and opposite in direction, resulting in no net flow of either gas across the system. The value of 𝑁𝐴𝑁𝐴+𝑁𝐵NA+NBNA can be calculated as: 𝑁𝐴𝑁𝐴+𝑁𝐵=𝑁𝐴𝑁𝐴−𝑁𝐴=10NA+NBNA=NA−NANA=01 However, division by zero is undefined. In the case of steady-state equimolal counter-diffusion, the molar flux of one gas is equal in magnitude but opposite in direction to the molar flux of the other gas. Therefore, the correct answer would be: 𝑁𝐴𝑁𝐴+𝑁𝐵=11+1=12=0.5NA+NBNA=1+11=21=0.5 So, the correct option is 0.5. |


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