Rates of Effusion: Homework Questions & Solutions

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In summary, Joe is trying to determine the atomic mass of an unknown gas, but gets confused when it is shown both ways and ends up getting the wrong answer.
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Homework Statement


So this isn't just one problem but it is a question as to why it is done one way for two instances and then the opposite way for a third problem. Here it goes.

1) An unknown gas effuses at a rate that is .462 times that of nitrogen gas (at the same temperature). Calculate the molar mass of the unknown gas in g/mol.

2) Uranium-235 can be separated from U-238 by flourinating the uranium to form UF6 (which is a gas) and then taking advantage of the different effusion rates for compounds containing the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6.

3) A sample of neon effuses from a container in 76 seconds. The same amount of an unknown noble gas requires 155 seconds. Identify the gas.

Homework Equations



[tex]\frac{RateA}{RateB}[/tex] = [tex]\sqrt{\frac{M_B}{M_A}}[/tex]

Where M = molar mass

The Attempt at a Solution



1) [tex]\frac{RateA}{RateB}[/tex] = .462

MA = [tex]\frac{M_B}{(.462)^2}[/tex]

MA = [tex]\frac{28.02}{(.462)^2}[/tex] = 131g


2) Rate A/ MA = U-238 RateB/ MB = U-235

[tex]\frac{RateA}{RateB}[/tex] = [tex]\sqrt{\frac{M_B}{M_A}}[/tex]

[tex]\sqrt{\frac{235.054}{238.051}}[/tex]= .9934

3) This is the one where I am confused on why it was carried out in the way as follows:
If I was to follow the format as shown above, where rateA is over rateB and then on the other side they switch positions, I arrive at a terribly incorrect answer:

[tex]\frac{76s}{155s}[/tex] = [tex]\sqrt{\frac{M_U}{20.18g}}[/tex]

.2404=[tex]\frac{M_U}{20.18}[/tex][tex]\Rightarrow[/tex] 4.85g which is the incorrect answer.

The only way I can arrive at the correct answer is when rate A is in the same postion on both sides for example:

[tex]\frac{76s}{155s}[/tex] = [tex]\sqrt{\frac{20.18g}{M_U}}[/tex]

MU = [tex]\frac{20.18g}{.2404}[/tex][tex]\Rightarrow[/tex] 83.9g

Which points to Krypton and that happens to be the correct answer. I realize this is a long question and I have the answer, but I want to make sure I am not missing something in the formatting and why it appears to change between problems. Thanks in advance.

Joe
 
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  • #2
You have just got momentarily confused into thinking the bigger number meant faster, but as it is time taken the bigger number means slower, so in your you first attempt you put it upside down. The ratio of atomic masses is close to 4 and Kr is clearly the answer they want.

(You might object that "effuses from a container in 76 seconds" is a pretty vague statement, but whatever it is it is the same for both and Kr is the answer they want. :biggrin:)
 

FAQ: Rates of Effusion: Homework Questions & Solutions

1. What is the definition of effusion?

Effusion is the process of a gas escaping through a small opening into a vacuum or another gas. It is also known as the flow of gas from one compartment to another.

2. How is the rate of effusion calculated?

The rate of effusion is calculated using Graham's Law, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula is expressed as r1/r2 = √(M2/M1), where r1 and r2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.

3. What factors affect the rate of effusion?

The rate of effusion is affected by several factors, including the molar mass of the gas, the temperature, the pressure, and the size of the opening through which the gas is effusing. Generally, lighter gases and higher temperatures result in faster rates of effusion.

4. How does effusion differ from diffusion?

While both effusion and diffusion involve the movement of gases from one area to another, they differ in their mechanisms. Effusion occurs when a gas escapes through a small opening, while diffusion involves the random movement of gas particles from an area of high concentration to an area of low concentration.

5. How is the concept of effusion applied in real life?

The concept of effusion is applied in various real-life scenarios, such as in the production of helium balloons, where helium gas is inserted into a balloon and escapes through the small pores in the balloon's material. It is also used in the manufacturing of semiconductors, where gases are effused through tiny holes to deposit thin films on electronic devices.

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