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Chapter 6: Problem 83

What are the ratios of the diffusion rates for the pairs of gases (a) \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{D}_{2} \mathrm{O}\) \(\left(\mathrm{D}=\text { deuterium, i.e., }_{1}^{2} \mathrm{H}\right) ;\) (c) \(^{14} \mathrm{CO}_{2}\) and \(^{12} \mathrm{CO}_{2}\) (d) \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6} ?\)

Short Answer

Expert verified
The ratios of the diffusion rates are: (a) 1.069 for N2:O2, (b) 1.054 for H2O:D2O, (c) 1 for ^14C:^12C, and (d) 1 for ^235U:^238U.

Step by step solution

01

Identify molecular weights

Calculate the relative atomic masses for all the given gases. They are calculated as N2 (28 g/mol), O2 (32 g/mol), H2O (18 g/mol), D2O (20 g/mol), ^{14}CO2 (44 g/mol), ^{12}CO2 (44 g/mol), ^{235}UF6 (352 g/mol) and ^{238}UF6 (352 g/mol).
02

Use Graham's Law

Apply Graham's law of diffusion, which states that the rate of diffusion \(r\) of a gas is inversely proportional to the square root of its molar mass \(M\). It is expressed as \( r_a / r_b = \sqrt{M_b / M_a} \) where \(r_a\) and \(r_b\) are the rates of diffusion, and \(M_a\) and \(M_b\) are the molar masses.
03

Calculate diffusion rates

Substitute the values of molar masses in Graham's law equation and compute the rates of diffusion for each pair: (a) N2 and O2: \( r_{N2} / r_{O2} = \sqrt{32g/mol / 28g/mol} = 1.069 \) (b) H2O and D2O: \( r_{H2O} / r_{D2O} = \sqrt{20g/mol / 18g/mol} = 1.054 \) (c) ^{14}CO2 and ^{12}CO2: \( r_{^14CO2} / r_{^12CO2} = \sqrt{44g/mol / 44g/mol} = 1 \) (d) ^{235}UF6 and ^{238}UF6: \( r_{^235UF6} / r_{^238UF6} = \sqrt{352g/mol / 352g/mol} = 1 \)

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