To determine which gas you would detect first on the other side of a large room, we need to consider the molecular properties of each gas, specifically their rates of diffusion. The rate of diffusion of a gas is inversely proportional to the square root of its molar mass, as described by Graham's law of effusion:
[tex]\[ \text{Rate of diffusion} \propto \frac{1}{\sqrt{M}} \][/tex]
Where [tex]\( M \)[/tex] is the molar mass of the gas.
1. Determine the molar masses of the gases:
- Hydrogen sulfide ([tex]\(H_2S\)[/tex]):
- Hydrogen ([tex]\(H\)[/tex]) has a molar mass of approximately 1 g/mol.
- Sulfur ([tex]\(S\)[/tex]) has a molar mass of approximately 32 g/mol.
- Total molar mass of [tex]\(H_2S\)[/tex]:
[tex]\[ M_{H_2S} = 2(1) + 32 = 34 \text{ g/mol} \][/tex]
- Ammonia ([tex]\(NH_3\)[/tex]):
- Nitrogen ([tex]\(N\)[/tex]) has a molar mass of approximately 14 g/mol.
- Hydrogen ([tex]\(H\)[/tex]) has a molar mass of approximately 1 g/mol.
- Total molar mass of [tex]\(NH_3\)[/tex]:
[tex]\[ M_{NH_3} = 14 + 3(1) = 17 \text{ g/mol} \][/tex]
2. Compare the rates of diffusion using Graham’s Law:
The gas with the smaller molar mass will diffuse faster. Comparing the molar masses of [tex]\( NH_3 \)[/tex] and [tex]\( H_2S \)[/tex]:
- [tex]\( M_{H_2S} = 34 \text{ g/mol} \)[/tex]
- [tex]\( M_{NH_3} = 17 \text{ g/mol} \)[/tex]
Since ammonia ([tex]\( NH_3 \)[/tex]) has a smaller molar mass than hydrogen sulfide ([tex]\( H_2S \)[/tex]), it will diffuse faster.
3. Conclusion:
- Between [tex]\( H_2S \)[/tex] and [tex]\( NH_3 \)[/tex], you would detect the odor of [tex]\( NH_3 \)[/tex] first on the other side of the room because ammonia has a smaller molar mass and thus diffuses more quickly compared to hydrogen sulfide.
So, the odor you would detect first on the other side of the room is [tex]\( NH_3 \)[/tex].
