Question #110935
Equal volumes of oxygen and nitrogen were allowed to diffuse under the similar conditions of temperature and pressure.If oxygen took 18s to diffuse , determine the time that would be taken by nitrogen
1
Expert's answer
2020-04-20T15:06:14-0400

According to Graham's law of diffusion, the ratio of the rates of diffusion of two gases (rO2r_{O_2},rN2r_{N_2}) is the square root of the inverse ratio of their molar masses (MO2M_{O_2} , MN2M_{N_2} ):

rO2rN2=MN2MO2\frac{r_{O_2}}{r_{N_2}} = \sqrt\frac{M_{N_2}}{M_{O_2}} .

The rate of diffusion is the amount of gas passed through an area per unit time. Therefore, it is inversely proportional to the time of diffusion tt :

rO2rN2=tN2tO2\frac{r_{O_2}}{r_{N_2}} =\frac{t_{N_2}}{t_{O_2}} .

Combine the equations above:

MN2MO2=tN2tO2\sqrt\frac{M_{N_2}}{M_{O_2}}=\frac{t_{N_2}}{t_{O_2}} .

The molar masses of O2 and N2 are 32.00 g/mol and 28.01 g/mol, respectively. The time taken by nitrogen is:

tN2=tO2MN2MO2=18(s)28.01 (g/mol)32.00 (g/mol)=16.8(s)t_{N_2}=t_{O_2}\sqrt\frac{M_{N_2}}{M_{O_2}} = 18(\text{s})\cdot \sqrt\frac{28.01\text{ (g/mol)}}{32.00\text{ (g/mol)}} = 16.8 (\text{s})

Answer: if oxygen took 18s to diffuse , the time that would be taken by nitrogen is 16.8s.


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