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Zorluk: OrtaGraham's Law of Diffusion and Effusion

Under fixed laboratory conditions of temperature and pressure, Gas A has a density of 0.09 g/dm30.09\text{ g/dm}^3 while Gas B has a density of 1.44 g/dm31.44\text{ g/dm}^3. What is the ratio of the rate of diffusion of Gas A to the rate of diffusion of Gas B?

  1. 4.04.0Cevap
  2. B
    16.016.0
  3. C
    0.250.25
  4. D
    0.06250.0625

Cevap

The ratio of the rate of diffusion of Gas A to Gas B is 4.04.0.
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its density or molar mass. Calculating 1.440.09=16=4.0\sqrt{\frac{1.44}{0.09}} = \sqrt{16} = 4.0 correctly gives the relative rate ratio, showing that the lighter gas diffuses 4 times faster than the denser gas.

Adım Adım Çözüm

1
State Graham's Law of Diffusion in terms of gas densities.
rArB=dBdA\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}
Graham's law dictates that the rate of diffusion (rr) of a gas is inversely proportional to the square root of its density (dd) at constant temperature and pressure.
2
Substitute the given density values into the equation.
rArB=1.44 g/dm30.09 g/dm3\frac{r_A}{r_B} = \sqrt{\frac{1.44\text{ g/dm}^3}{0.09\text{ g/dm}^3}}
Gas B's density (dB=1.44d_B = 1.44) goes into the numerator and Gas A's density (dA=0.09d_A = 0.09) into the denominator due to the inverse relationship.
3
Simplify the fraction inside the square root and calculate the final square root.
rArB=16=4.0\frac{r_A}{r_B} = \sqrt{16} = 4.0
Dividing 1.441.44 by 0.090.09 gives 1616, and taking the square root of 1616 yields 4.04.0.

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Graham's Law relating gas diffusion rates to density
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Graham's Law of Diffusion and Effusion

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