Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Effusion involves a gas escaping a container through an opening.
Concept: Understanding Graham's Law of Effusion.3m
Welcome back guys. In this new video, we'll get to look at the effusion rates of different gases.
Tied with effusion is basically the concept of velocity and speed. Remember we use root mean square equations when we're dealing with the speed of one gas. We use effusion when we're comparing the speeds or velocities of multiple gases. That's when we use this concept.
Here we're going to say Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. What this is really saying is that the speed or rate at which a gas moves, so effusion is just seen as speed, the speed of a gas is the opposite of its weight.
What that means is if my speed is high, it's because the mass of my gas is low. Basically, the less you weigh as a gas, the faster you move. That's what effusion is trying to explain. Effusion looks and compares multiple gases to each other. Gas A moves faster than Gas B. Why is that? Well, Gas A must weigh less. That's why it's moving faster.
We're going to say when we're comparing the rates of two gases, we use this form of Graham's Law of Effusion. Now pay close attention. Basically, we're going to say the rate of Gas One is equal to its rate divided by the rate of Gas Two. We're going to say that equals the square root – look if Gas One's rate is on the top, then it's mass is on the bottom because remember Graham's Law says that rate and mass are inversely proportional. They're on opposite levels. Here if the rate of Gas Two is on the bottom, then its mass is on the top. We're going to have to take this approach to answer some of these questions.
When comparing the rate or speed of two gases then we must use Graham’s Law of Effusion.
Example: Calculate the ratio of the effusion rates of helium and methane (CH4).4m
The gas named first will represent gas 1 and the second gas will represent gas 2. Once that is established we simply use Graham’s Law of Effusion.
Example: Rank the following in order of increasing rate of effusion:
O2 AlF5 CO2 Xe3m
How much faster do ammonia (NH 3) molecules effuse than carbon monoxide (CO) molecules?
Enter the ratio of the rates of effusion. Express your answer numerically using three significant figures.
An unknown gas Q diffuses at a RATE 1.65 times faster than that of propane, C 3H8. Which of these are most likely to be Q?
Given the gases Ne, He, H2, Kr, put them in order of their INCREASING rate of effusion.
1. Kr < Ne < He < H 2
2. H2 < He < Ne < Kr
3. He < H2 < Kr < Ne
4. Ne < Kr < H 2 < He
At a given temperature and pressure, a sample of Gas A is observed to diffuse twice as fast as a sample of a different gas, B. Based on this:
A. The molar mass of A is one fourth that of B
B. The molar mass of A is one half that of B
C. The molar mass of A is 0.707 times that of B
D. The molar mass of A is 1.414 times that of B
E. The molar mass of A is four times that of B
The gas for which instance will effuse faster?
Second-hand smoke consist of many molecules including nicotine. How much faster or slower does the nicotine diffuse, compared to carbon dioxide? (MW nicotine = 160g/mol)
1. About half as fast
2. As gases, the two molecules diffuse at the same rate
3. About 4 times faster
4. About a fourth as fast
5. About twice as fast
Oxygen gas effuses at a rate that is 1.49 times that of an unknown gas (at the same temperature). What is the identity of the unknown gas?
NO2 gas effuses ________times faster than Cl 2 gas.
Consider the chemical reaction: 2 SO2 (g) → SO(g) + SO3 (g)
Determine if the sentence below is true or false
Between the two products, SO3 would have the highest rate of effusion.
The mass particle is 10% larger than that of particle B. If particle B travels 100 m during effusion, how far would particle A travel during effusion?
a) 90 m
b) 95 m
c) 100 m
d) 105 m
e) 110 m
Using Graham's Law of effusion, calculate the ratio of the effusion rates of H 2 to CO through a pinhole.
How many times faster will He gas pass through a pin hole than SO2 gas?
e) none of the given answers
Carbon dioxide effuses through a pinhole at a rate of 0.232 ml/min at 25.0 °C. Another gas effuses at a rate of 0.363 ml/min. What is the molar mass of the gas?
At any given temperature, how much more quickly will He effuse than Xe?
1. 5.7 times more quickly
2. .03 times more quickly
3. 8.1 times more quickly
4. They will diffuse at the same rate
5. 0.12 times more quickly
6. 32.8 times more quickly
A sample of helium effuses through a porous container 7.70 times faster than does unknown gas X. What is the molar mass of the unknown gas?
At a particular temperature and pressure, it takes 4.55 min for a 1.5 L sample of He to effuse through a porous membrane. How long would it take for a 1.5 L of F2 to effuse under the same conditions?
a. 18 min
b. 14 min
c. 10 min
d. 5 min
e. 1.5 min
The mass of particle A is 10% larger than that of particle B. If particle B travels 100m during effusion, how far would particle A travel during effusion?
a. 90 m
b. 95 m
c. 100 m
d. 105 m
e. 110 m
A sample of N2 gas is contaminated with a gas (A) of unknown molar mass. The partial pressure of each gas is known to be 200 Torr at 25°C. The gases are allowed to effuse through a pinhole, and it is found that gas A escapes at three times the rate of N2. The molar mass of gas A is:
e) none of these
Exactly 10 mL, of gas A (MM = 15.0 g/mol) effuse through an opening in 2.0 sec. Exactly 10 mL of gas B effuse through the same opening in 8.0 sec. What is the molar mass of gas B?
a) 60 g/mol
b) 480 g/mol
c) 240 g/mol
d) 120 g/mol