# a vessel which contains gases a and b at 1 atm and 4.5 c is connected to another vessel with different concentration of gases at the same temperature and pressure. the vessels are connected by a 23 cm long tube with inside diameter of 3.8 cm at the tube with inside diameter of 3.8 cm at the tube inlet and of 2.5 cm at the tube outlet. determine steady state rate of transport of gas a between the vessels if concentration of a in one tank is 85% mole and that in the other thank is 7% mole. transfer occurs by molecular diffusion and the diffusity is dab = 4.5 x 10^-5 m/sec.

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a vessel which contains gases a and b at 1 atm and 4.5 c is connected to another vessel with different concentration of gases at the same temperature and pressure. the vessels are connected by a 23 cm long tube with inside diameter of 3.8 cm at the tube with inside diameter of 3.8 cm at the tube inlet and of 2.5 cm at the tube outlet. determine steady state rate of transport of gas a between the vessels if concentration of a in one tank is 85% mole and that in the other thank is 7% mole. transfer occurs by molecular diffusion and the diffusity is dab = 4.5 x 10^-5 m/sec.

1st mistake:units for diffusivity Dab are m^2/sec.

2nd: What is 4.5 c i didn’t get that

3rd:transfer occurs by diffusion is ok ,but it is necessary to provide whether this diffusion is taking place by either diffusion of gas a by stagnent gas b or equimolal counter diffusion.ok,nelect first two but what abot the type of diffusion taking place?

Ok,here I am assuming the diffusion taking place by gas a through stagnent(non diffusing) gas b,then

the rate of transport is given by Wa=(Na)*(S)

where,Wa=amount of rate of component transfer[kilo mole/s or kg/sec],

Na=flux[kilo mole/(m^2)-sec]

S=surface area(m^2) we should consider diameter at the exit of 1st pipe or entrance of 2nd tube which are same and equal to 2.5cm.