To determine which gas you would detect first on the other side of a room when both [tex]\( H_2S \)[/tex] and [tex]\( NH_3 \)[/tex] are released simultaneously, we can use Graham's Law of Effusion. This law states that the rate of diffusion of a gas (or its velocity of effusion) is inversely proportional to the square root of its molar mass. Symbolically, this can be expressed as:
[tex]\[ \text{Rate} \propto \frac{1}{\sqrt{\text{Molar Mass}}} \][/tex]
Now, let's consider the molar masses of the two gases:
- [tex]\( H_2S \)[/tex] has a molar mass of 34.08 grams per mole.
- [tex]\( NH_3 \)[/tex] has a molar mass of 17.031 grams per mole.
According to Graham’s Law, the gas with the lower molar mass will diffuse faster. In this case, [tex]\( NH_3 \)[/tex] has a lower molar mass compared to [tex]\( H_2S \)[/tex].
To provide a precise comparative measure of their diffusion rates, we express the rates as follows:
1. Calculate the diffusion speeds inversely proportional to the square root of their molar masses:
[tex]\[
\text{Diffusion speed of } H_2S = \frac{1}{\sqrt{34.08}} \approx 0.1713
\][/tex]
[tex]\[
\text{Diffusion speed of } NH_3 = \frac{1}{\sqrt{17.031}} \approx 0.2423
\][/tex]
Given these diffusion speeds:
- The diffusion speed of [tex]\( NH_3 \)[/tex] is approximately 0.2423.
- The diffusion speed of [tex]\( H_2S \)[/tex] is approximately 0.1713.
Since 0.2423 (the diffusion speed of [tex]\( NH_3 \)[/tex]) is greater than 0.1713 (the diffusion speed of [tex]\( H_2S \)[/tex]), [tex]\( NH_3 \)[/tex] will diffuse faster across the room.
Therefore, you would detect the odor of [tex]\( NH_3 \)[/tex] first on the other side of the room.
