Graham's Law of Effusion
Problems 11-25

Probs 1-10

Ten Examples

Examples and Problems only

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Problem #11: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.62 mol/min?

Solution:

1) Set rates:

rate₁ = 3.62
rate₂ = x

2) The gas that has twice the molar mass is the one whose rate we are trying to determine.

MM₂ = 2
MM₁ = 1

These molar masses are arbitrary values, we just need MM₂ to be twice the value for MM₁.

3) Graham's Law:

r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

3.62 / x = $\sqrt{\mathrm{2\; /\; 1}}$

x = 2.56 mol/min

Problem #12: Calculate the rate of effusion of NO₂ compared to SO₂ at the same temperature and pressure.

Solution:

1) The rates of effusion of two gases are inversely proportional to the square roots of their molar masses. Here is a statement of Graham's law:

rate₁² x MM₁ = rate₂² x MM₂

2) Solve for the unknown:

rate₂ = $\sqrt{\mathrm{(rate₁²\; x\; MM₁)\; /\; MM₂}}$

rate₂ = $\sqrt{\mathrm{(1²\; x\; 46.005)\; /\; 64.063}}$

rate₂ = 0.85

The rate of effusion of SO₂ is 0.85 times the rate of effusion of NO₂, which is logical because SO₂ is more massive than NO₂, and moves more slowly, on average.

3) The above solution was not written the ChemTeam, but it is rather nice, so I decided to copy it as is. Notice how the solution assigns rate₁ to be equal to 1. You might wish to rearrange the writer's Graham's law equation into the one the ChemTeam tends to use.

Problem #13: Assume you have a sample of hydrogen gas containing H₂, HD, and D₂ that you want to separate into pure components. What are the various ratios of relative rates of effusion?

Solution:

Let us first compare H₂ and HD to D₂. Since D₂ is the heaviest molecule, it is the slowest. D₂'s rate (which is r₂) will be set to 1.

Graham's Law is: r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

1) H₂ : D₂

x / 1 = $\sqrt{\mathrm{4.028\; /\; 2.016}}$

x = 1.4135

H₂ effuses 1.4135 times faster than D₂

2) HD : D₂

x / 1 = $\sqrt{\mathrm{4.028\; /\; 3.022}}$

x = 1.1545

HD effuses 1.1545 times faster than D₂

3) Finally, let us compare H₂ to HD. This may be solved two different ways:

x / 1 = $\sqrt{\mathrm{3.022\; /\; 2.016}}$

x = 1.2243

or, use a ratio and proportion:

1.4135 is to 1.1545 as x is to one

x = 1.2243

H₂ effuses 1.2243 times faster than HD.

Problem #14: A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse though the membrane?

Solution:

1) Determine helium's rate of effusion:

1.50 L per 25 hr = 0.0600 L/hr.

Let r₂ be the rate for helium.

That means r₁ will be the rate for oxygen in L/hr.

2) Determine oxygen's rate of effusion:

r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 0.0600 = $\sqrt{\mathrm{4.0026\; /\; 31.9988}}$

x = 0.0212205 L/hr

3) Determine time for half of oxygen's 3.00 liters to effuse:

1.50 L / 0.0212205 L/hr = 70.67 hrs

Problem #15: At a certain temperature, hydrogen molecules move at an average velocity of 1.84 x 10³ m/s. Estimate the molar mass of a gas whose molecules have an average velocity of 311 m/s.

Solution:

1840 / 311 = $\sqrt{\mathrm{x\; /\; 2.016}}$

Divide, square both sides, multiply by 2.016

x = 70.57 g/mol

Although the question does not ask for the identity of the gas, we could identify it tentatively (based on just the data we have) as Cl₂. The molecular weight for chlorine gas is 70.9 g/mol.

Problem #16: An unknown gas effuses 1.66 times more rapidly than CO₂. What is the molar mass of the unknown gas.

Solution:

1 / 1.66 = $\sqrt{\mathrm{x\; /\; 44.009}}$

Divide, square both sides, multiply by 44.009

x = 16.0 g/mol

Although the question does not ask for the identity of the gas, we could identify it tentatively (based on just the data we have) as CH₄. The molecular weight for methane gas is 16.043 g/mol.

Problem #17: A sample of hydrogen gas effuse through a porous container 8.91 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Solution:

9 / 1 = $\sqrt{\mathrm{x\; /\; 2.016}}$

x = 160.05 g/mol

Based only on the molecular weight, the unknown gas could be bromine.

Problem #18: N₂ is contaminated with a noble gas.The contaminant effuses at 1.87x N₂. What is the noble gas?

Solution:

r₁ = N₂ = 1
r₂ = unk gas = 1.87

r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

1/1.87 = $\sqrt{\mathrm{x\; /\; 28.014}}$

x = 8.011 g/mol

No noble gas has molar mass of 8.011 g/mol.

However, He = 4.0026, so perhaps the desired answer is He₂. The problem, of course, is that He₂ does not exist.

The ChemTeam's personal opinion is that the writer (NOT the ChemTeam!) used 14.007 rather than 28.014 in designing the problem. Using 14.007 gives an answer of 4.0055, very close to the atomic weight of He.

Problem #19: In an effusion experiment, it was determined that nitrogen gas, N₂, effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?

Solution:

(r₁/r₂)² = MM₂/MM₁

Notice the slightly different formulation of Graham's Law.

r₁ = N₂ = 1.812
r₂ = unk gas = 1

(1.812 / 1)² = (x / 28.014)

x = 91.98 g/mol

Problem #20: Why are the rates of diffusion of nitrogen gas and carbon monoxide almost identical at the same temperature?

Solution:

1) The speed of a gas is given by:

v = $\sqrt{\mathrm{3RT\; /\; M}}$

where M is the molecular weight of the gas in kg/mol.

2) The molecular weights are:

N₂ = 0.028014 kg/mol
CO = 0.028010 kg/mol

You should be able to see that the speeds will be nearly identical without solving the formula. Two values (R and T) are going to be same for each gas and the values for M are very nearly the same, differing by only 0.000004.

The diffusion rates for nitrogen gas and carbon monoxide gas are very nearly the same at the same temperature because the two substances have very nearly the same molecular weights.

Problem #21: In running a diffusion experiment, ammonia is found to diffuse 30.0 cm during the same amount of time hydrogen chloride moves 20.0 cm. Calculate the percentage deviation from Graham's Law.

Solution:

The experimentally determined ratio (ammonia divided by HCl) is 1.50.

1) What ratio is predicted by Graham's Law:

r₁ = NH₃ = x
r₂ = HCl = 1

MM₁ = 17.0307 g/mol
MM₂ = 36.4609 g/mol

x / 1 = $\sqrt{\mathrm{36.4609\; /\; 17.0307}}$

x = 1.463

Ammonia diffuses 1.463 times faster than HCl.

2) Percent deviation is:

(1.50 − 1.463) / 1.50 = 2.47%

Problem #22: A sample of Br₂(g) take 10.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

Solution:

Let us assume that 1.00 mole of Br₂ effuses. Therefore, its rate is 1.00 mol / 10.0 min = 0.100 mol/min

r₁ = x
r₂ = 0.100 mol/min

MM₁ = 39.948 g/mol
MM₂ = 159.808 g/mol

x / 0.100 = $\sqrt{\mathrm{159.808\; /\; 39.948}}$

x / 0.100 = 2.00

x = 0.200 mol/min

1.00 mole of Ar effuses in 5.00 minutes

Problem #23: At a particular pressure and temperature, it takes just 8.256 min for a 4.893 L sample of Ne to effuse through a porous membrane. How long would it take for the same volume of I₂ to effuse under the same conditions?

Solution:

r₁ = x
r₂ = 4.893 L / 8.256 min = 0.59266 L/min

MM₁ = 253.8 g/mol
MM₂ = 20.18 g/mol

x /0.59266 = $\sqrt{\mathrm{20.18\; /\; 253.6}}$

x / 0.59266 = 0.2821

x = 0.16719 L/min

4.893 L / 0.16719 L/min = 29.27 min

Problem #24a: How much faster does U²³⁵F₆ effuse than U²³⁸F₆?

Solution:

1) Calculate molecular weights:

U²³⁵F₆ = 235.04393 + 6(18.99840) = 349.03433

U²³⁸F₆ = 238.05079 + 6(18.99840) = 352.04119

2) U²³⁸F₆ is heavier, so:

assign its rate to r₂ and set the rate equal to 1

3) Solve Graham's Law:

r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 1 = $\sqrt{\mathrm{352.04119\; /\; 349.03433}}$

x = 1.0043

U²³⁵F₆ effuses 1.0043 times faster than U²³⁸F₆

The following problem is worded so as to use the exact reverse ratio in problem 24a.

Problem #24b: Calculate the ratio of effusion rates for U²³⁸F₆ and U²³⁵F₆. Express your answer using five significant figures and as the following ratio:

rate U²³⁸F₆ / rate U²³⁵F₆

Solution:

Specifying the form of the ratio forces the numbers to be placed in certain places in Graham's law The U-238 values MUST be associated with r₁ and MM₁. The U-235 values MUST be associated with r₂ and MM₂:

r₁/ r₂ = $\sqrt{\mathrm{MM₂\; /\; MM₁}}$

x / 1 = $\sqrt{\mathrm{349.03433\; /\; 352.04119}}$

x = 0.99572

Problem #25: O₃ effuses 0.8165 times as fast as O₂. What % of the molecules effusing first would be O₂?

Solution:

The rate of effusion of O₂ is 1.225 times faster than O₃, which means that every second there will be 1225 molecules of O₂ effusing for every 1000 molecules of O₃. Therefore, the percentage of O₃ molecules is:

[1225 / (1225 + 1000)] x 100 = 55%

Bonus Problem #1: HCl and NH₃ diffuse through a tube and a white disc of NH₄Cl is formed. Where in the tube?

Solution:

1) Write Graham's Law:

r₁ / r₂ = √[MM₂ / MM₁]

2) Some comments:

(a) assign r₂ to be a value of 1.000
(b) Assign r₂ to be HCl (the heavier of the two)

3) Solve Graham's Law for r₁:

r₁ / 1.00 = √[36.4609 / 17.0307]

r₁ = 1.463

4) More comments:

(a) Those numbers mean that, for every amount the HCl (slower because it weighs more) moves, the NH₃ (faster because it weighs less) will move 1.463 times that amount.
(b) If we were to specify a length of tube, we could determine where the disc would form.

5) An example:

Set tube length to be 2.500 meters.

x + 1.463x = 2.500

x = 1.015 m <--- the distance the HCl moves

(1.463) (1.015) = 1.485 m <--- the distance the NH₃ moves

Bonus Problem #2: One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of:

¹²C¹⁶O
¹²C¹⁷O
¹²C¹⁸O

Solution:

1) First, the molar masses are these:

¹²C¹⁶O = 28.00
¹²C¹⁷O = 29.00
¹²C¹⁸O = 30.00

2) Assign a relative rate of 1.000 to ¹²C¹⁸O:

The ¹⁸O form of CO is picked because it is the heaviest, therefore the slowest. The other two will have a relative rate slightly greater than 1.

This is done purely for convenience. Any of the three forms of CO could be assigned a rate of 1.000.

3) Compare ¹²C¹⁷O to ¹²C¹⁸O:

r₁ / r₂ = √(MM₂ / MM₁)

x / 1 = √(30/29)

x = 1.017

4) Compare ¹²C¹⁶O to ¹²C¹⁸O:

r₁ / r₂ = √(MM₂ / MM₁)

x / 1 = √(30/28)

x = 1.035

You may wish to ponder this: name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

Probs 1-10

Ten Examples

Examples and Problems only

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