To determine which of the following gases diffuses the fastest — [tex]\(O_2\)[/tex], [tex]\(CH_4\)[/tex], [tex]\(CO_2\)[/tex], and [tex]\(Cl_2\)[/tex] — we can use Graham's Law of Diffusion. Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The formula for the rate of diffusion ([tex]\(r\)[/tex]) is given by:
[tex]\[ r \propto \frac{1}{\sqrt{M}} \][/tex]
where [tex]\(M\)[/tex] is the molar mass of the gas.
Let's break down the solution step by step.
1. List the molar masses of the gases:
- [tex]\(O_2\)[/tex]: 32 g/mol
- [tex]\(CH_4\)[/tex]: 16 g/mol
- [tex]\(CO_2\)[/tex]: 44 g/mol
- [tex]\(Cl_2\)[/tex]: 71 g/mol
2. Calculate the inverse of the square root of each molar mass:
- For [tex]\(O_2\)[/tex]:
[tex]\[
\text{Rate}_\text{O2} = \frac{1}{\sqrt{32}} \approx 0.1768 \text{ (rounded to 4 significant figures)}
\][/tex]
- For [tex]\(CH_4\)[/tex]:
[tex]\[
\text{Rate}_\text{CH4} = \frac{1}{\sqrt{16}} = 0.25
\][/tex]
- For [tex]\(CO_2\)[/tex]:
[tex]\[
\text{Rate}_\text{CO2} = \frac{1}{\sqrt{44}} \approx 0.1508 \text{ (rounded to 4 significant figures)}
\][/tex]
- For [tex]\(Cl_2\)[/tex]:
[tex]\[
\text{Rate}_\text{Cl2} = \frac{1}{\sqrt{71}} \approx 0.1187 \text{ (rounded to 4 significant figures)}
\][/tex]
3. Compare the rates of diffusion:
- Rate of [tex]\(O_2\)[/tex]: 0.1768
- Rate of [tex]\(CH_4\)[/tex]: 0.25
- Rate of [tex]\(CO_2\)[/tex]: 0.1508
- Rate of [tex]\(Cl_2\)[/tex]: 0.1187
According to these calculated rates, [tex]\(CH_4\)[/tex] has the highest rate of diffusion because 0.25 is the largest value among the rates. Thus, [tex]\(CH_4\)[/tex] diffuses the fastest among the given gases.
Answer: [tex]\(CH_4\)[/tex]
