c) This relationship is known as Graham's Law of Effusion. Since both gases are at te same temperature, they must have the same average kinetic energy ( mv2), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic H H LL H H energy, you can state that % MLVL2 = v2. Multiplying both sides by 2 gives you m v 2 y2. Rearranging the equation to = m both masses on the same side of the equation will give you m/mH = V L2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, V 2VL2 = 4. How many times heavier is the heavy gas compared to the light gas? d) If the light gas was Ne, what would be a reasonable identity for the heavy gas?

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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c) This relationship is known as Graham's Law of Effusion. Since both gases are at te same temperature, they must
have the same average kinetic energy (½ mv²), where m is mass and v is velocity (like speed). Since both gases have
the same average kinetic
energy, you can state that ½ muvL2 = v ². Multiplying both sides by 2 gives you m v 2 y ². Rearranging the equation to get
H H
LL
H H
2 m
= m
both masses on the same side of the equation will give you mu/mH = V 2/VL2. In 3a and 3b, you probably noticed that the
heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are
moving twice as fast, VH/VL = ½. Therefore, V 2/VL2 = ¼. How many times heavier is the heavy gas compared to the
light gas?
d) If the light gas was Ne, what would be a reasonable identity for the heavy gas?
Transcribed Image Text:c) This relationship is known as Graham's Law of Effusion. Since both gases are at te same temperature, they must have the same average kinetic energy (½ mv²), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ muvL2 = v ². Multiplying both sides by 2 gives you m v 2 y ². Rearranging the equation to get H H LL H H 2 m = m both masses on the same side of the equation will give you mu/mH = V 2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, V 2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas? d) If the light gas was Ne, what would be a reasonable identity for the heavy gas?
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