A Sample Of N2 Effuses In 220 S. How Long Will The Same Size Sample Of Cl2 Take To Effuse?

Graham's Constabulary of Effusion
Ten Examples

Boyle's Law	No Name Law	Graham'southward Law Probs ane-10
Charles' Constabulary	Combined Gas Law	Graham's Police force Probs 11-25
Gay-Lussac's Law	Ideal Gas Constabulary	All Graham's Police examples and problems
Avogadro's Law	Dalton's Law	Return to KMT & Gas Laws Menu
Diver'south Law

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Discovered past Thomas Graham of Scotland former in the 1830'southward (the ChemTeam is non sure exactly when).

To discuss this law, please consider samples of two different gases at the same Kelvin temperature.

Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy (KE) of the two gas samples is also the same.

In equation form, we can write this:

KE_one = KE_two

Since KE = ^one⁄₂ mvⁱⁱ, (1000 = mass and v = velocity) we can write the following equation:

1000₁5₁ ⁱⁱ = m_iiv_two ²

Note that the value of one-half cancels.

The equation in a higher place can be rearranged algebraically into the post-obit:

$\sqrt{\mathrm{grand_one\; /\; m_ii}}$ = v₂ / v₁

Yous may wish to assure yourself of the definiteness of this rearrangement (likewise every bit the one only below).

Graham'due south Police force is ofttimes stated as follows:

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

where MM means the molar mass of the substance in question. Often, in these types of bug, you will exist called upon to determine the molar mass of an unknown gas.

Simply above, I formatted Graham's Constabulary using MathML. In a variety of places on the Internet, MathML cannot exist used. Expect for this:

r_i / r₂ = SQRT(MM_two / MM_i)

or this:

$\text{r}{}_{\mathrm{ane}}^{\mathrm{ii}}$ / $\text{r}{}_{\mathrm{ii}}^2$ = MM_two / MM₁

every bit alternate ways to write Graham's Constabulary.

There is also this manner:

r₁ / r_ii = √(MM₂ / MM₁)

This method uses the foursquare root symbol that is bachelor through HTML code.

Note that I have substituted r for v. In Graham'south Constabulary, we will look at the rate of effusion (movement of gas through a small-scale pinhole into a vacuum) more than ofttimes than nosotros will await at a speed (like a root mean square speed). That ways nosotros are mostly looking at amounts that move per unit fourth dimension, not how fast the individual particles are moving. For example, a charge per unit unit might be mL/min, which is not a unit of speed. Some other example would be moles per second.

Last betoken: ofttimes the rate of effusion of i gas is given relative to the rate of effusion of the other gas. That allows you to ready the rate of effusion for one of the gases to a numerical value of 1. This is employed oft, look for information technology.

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Example #ane: 8.278 x 10¯⁴ mol of an unidentified gaseous substance effuses through a tiny pigsty in 86.ix s Under identical conditions, 1.740 x 10¯⁴ mol of argon gas takes 81.3 s to effuse.

a) What is the molar mass of the unidentified substance (in g/mol)?
b) What is the molecular formula of the substance?
c) Under identical conditions, how many moles of ethene (C₂H_iv) gas would effuse in 91.0 s?

Solution:

1) Calculate the rates of effusion:

unknown ---> 8.278 x 10¯⁴ mol / 86.9 south = nine.525892 x 10¯⁶ mol/southward
argon ---> one.740 x ten¯⁴ / 81.3 s = 2.140221 x 10¯⁶ mol/s

Annotation that these are not speeds. A speed is an amount of distance covered in a unit of measurement amount of time. The to a higher place is a charge per unit, a number of moles of gas effuse through a pinhole in a unit of measurement amount of time.

two) Use Graham's Constabulary:

r_i / r₂ = $\sqrt{\mathrm{MM_two\; /\; MM_i}}$

Assign the unknown molar mass to be MM₂. I will cancel the exponent on the rates, since they are both 10¯⁶.

2.140221 / 9.525892 = $\sqrt{\mathrm{MM_two\; /\; 39.948}}$

3) Solve:

0.05047844 = MM₂ / 39.948

MM₂ = 2.0165 grand/mol (the reply to role a)

The gas is hydrogen, H₂ (the answer to part b; no other gas weighs 2).

4) The solution to part c:

the molecular weight of ethene is 28.0536 1000/mol

allow us use the data from argon

allow the rate for ethene exist r₁

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

x / 2.140221 x 10¯^{half dozen} = $\sqrt{\mathrm{39.948\; /\; 28.0536}}$

ten / 2.140221 ten 10¯⁶ = 1.19331

ten = 2.55395 ten ten¯⁶ mol/s

2.55395 x 10¯⁶ mol/south times 91.0 s = ii.324 x 10¯⁴ mol

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Example #2: Information technology takes 354 seconds for 1.00 mL of Xe to effuse through a pocket-sized hole. Under the same conditions, how long volition it accept for 1.00 mL of nitrogen to effuse?

Solution:

1) Write the rates for each gas:

xenon ---> i.00 mL / 354 due south
nitrogen ---> 1.00 mL / x

A common temptation is to use the times directly, every bit in Xe rate = 354 sec and N_two charge per unit = x. Still, delight remember that rates accept time in the denominator.

I'm going to leave the rates as fractions when entering them into Graham's Law. Since I knew what I was going to do when I wrote this solution, I'chiliad going to assign nitrogen to r₂.

2) Graham'due south Law:

r_ane / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM_i}}$

[(ane/354) / (1/x)] = $\sqrt{\mathrm{28.014\; /\; 131.293}}$

Watch what happens to the left-hand side.

x / 354 = 0.46192

x = 163.5 southward

Annotation that this is a reasonable answer. Since the nitrogen is lighter than xenon, it takes less time for 1.00 mL of nitrogen to effuse. If you lot had used the times in the numerator of the rate (equally opposed to being in the denominator, where they are supposed to exist), you lot would accept calculated an respond much larger than 354 s.

Note that the common mistake described tin exist fabricated past anybody! The ChemTeam made the above mistake on ane of the problems in one of the linked files and did not catch it until a student emailed with a question virtually the mistaken solution. The ChemTeam was thankful (and embarrassed) and quickly fixef the error and uploaded the new file.

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Example #3: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a charge per unit of 4.2 mol/min?

Solution:

1) Permit the states set up the molar masses:

gas A = 1
gas B = ii

I'yard going to assign gas B to r₁ (and MM₁, of grade) since we desire to know the effusion charge per unit for the gas that is twice the molar mass of the other. (Personal note: I similar putting the unknown into the numerator. It seems to fit better with my brain.)

Notice how I simply employ ane and two for the molar masses. We only know that one is twice the other, we do not know the actual values.

2) Employ Graham's Law:

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

10 / 4.2 = $\sqrt{\mathrm{one\; /\; 2}}$

x / four.2 = 0.70710678

x = two.97 mol/min

to ii significant figures, the reply is iii.0 mol/min

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Example #iv: Information technology takes 110. seconds for a sample of carbon dioxide to effuse through a porous plug and 275 seconds for the aforementioned book of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas (in g/mol)?

Solution:

1) Ready the rates, assuming 1.00 mole of each gas effuses:

r₁ = 1.00 mol / 110. s = 0.00909091 mol/southward
r₂ = i.00 mol / 275 s = 0.003636364 mol/south

Pease notation that I did non employ the times directly in the problem. I used them to create the rates. In footstep 5 beneath, I used the 2 fractions direct as opposed to what I did but above.

2) Set the tooth masses:

MM_one = 44.009 g/mol
MM₂ = x

3) Set up and substitute into Graham's Constabulary:

r_one / r_two = $\sqrt{\mathrm{MM_ii\; /\; MM₁}}$

0.00909091 / 0.003636364 = $\sqrt{\mathrm{x\; /\; 44.009}}$

4) Square both sides:

6.25 = 10 / 44.009

x = 275 thousand/mol

5) Another way:

[(1/110) / (ane/275)] = $\sqrt{\mathrm{x\; /\; 44.009}}$

275 / 110 = $\sqrt{\mathrm{x\; /\; 44.009}}$

2.5 = $\sqrt{\mathrm{ten\; /\; 44.009}}$

half dozen.25 = x / 44.009

x = 275 chiliad/mol

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Case #5: What is the tooth mass of a compound that takes 2.65 times as long to effuse through a porous plug equally it did for the same amount of XeF₂ at the same temperature and pressure?

Solution: Annotate: you lot have to be careful reading the problem, in order to keep the gases with the correct subscript. Note how I assign a rate of 1 to the XeF_ii. I exercise this considering the unknown gas takes 2.65 times longer than the XeF₂ to effuse.

1) Set the rates:

r₁ = two.65
r_two = ane

two) Set the molar masses:

MM₁ = ten
MM₂ = 169.286 g/mol

iii) Gear up and substitute into Graham's Police force:

r₁ / r_two = $\sqrt{\mathrm{MM_two\; /\; MM_one}}$

2.65 / one = $\sqrt{\mathrm{x\; /\; 169.286}}$

4) Square both sides:

vii.0225 = x / 169.286

10 = 1188.8 k/mol

Three significant figures seems best, so 1190 thousand/mol.

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Example #6: If a gas effuses iv.25 times faster than iodine gas (I₂), what is its molar mass?

(a) 59.7 thousand/mol

(b) 163 1000/mol

(c) 123 one thousand/mol

(d) 158 g/mol

Solution:

1) Graham's Constabulary:

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

2) We will set the rate for I₂ to be 1.00, which makes the charge per unit for the unknown gas exist iv.25. The molar mass of I₂ is 253.809 and the molar mass of the unknown is x

1 / 4.25 = $\sqrt{\mathrm{10\; /\; 253.809}}$

I assigned the r₂ and MM_two to the unknown gas so every bit to put x in the numerator.

three) Square both sides, then continue on to the value for x:

0.0553633 = x / 253.809

x = 14.0

four) I think that the virtually reasonable scenario is that the person who did the trouble to go the answer key did not foursquare the left-hand side of the equation. In other words, they just squared the right-mitt side of the equation. So, they wound upwards with this:

ane / iv.25 = x / 253.809 <--- I left the i/4.25 as a fraction rather than making information technology into a decimal value

And that leads to reply selection (a).

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Example #7: Summate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.threescore min and the same volume of nitrogen, at the aforementioned temperature and pressure, effuses through this apparatus in viii.50 minutes.

Solution:

1) Let us assume 1.00 L of each gas effused. This allows us to summate rates:

unknown ---> i.00 Fifty / 6.lx min = 0.1515152 L/min
Northward_two ---> 1.00 50 / viii.50 min = 0.117647 L/min

2) Now, nosotros tin use Graham's Law:

r₁ / r_ii = SQRT(MM₂ / MM₁)

I volition phone call the unknown r₂ considering that will put its molar mass in the numerator.

0.117647 / 0.1515152 = SQRT( x / 28.014)

0.881175 = x / 28.014

x = 24.685 g/mol

3) Since we know everything was at STP, the density is as follows:

24.685 k/mol / 22.414 L/mol = ane.10 thousand/L

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Case #8: If 0.0949 moles of NH₃ effuses in 881 seconds, how many seconds would information technology take for the same number of moles of B₂H₆ to effuse?

Solution:

1) Let us make up one's mind the charge per unit at which ammonia effuses:

0.0949 mol / 881 south = 0.0001077185 mol/s

2) Write Graham's Law:

r_one / r_two = SQRT(MM₂ / MM₁)

or

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

iii) Substitute values into Graham'southward Law and solve:

r₁ / 0.0001077185 mol/s = $\sqrt{\mathrm{17.0307\; /\; 27.6694}}$

r₁ / 0.0001077185 mol/s = 0.7845423

r₁ = 0.0000845097 mol/south

4) One terminal step:

0.0949 mol / 0.0000845097 mol/s = 1123 southward

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Example #9: The charge per unit of diffusion of an unknown gas was determined to be 2.92 times greater than that of NH_three. What is the approximate molar mass of the unknown gas?

Solution:

1) Write Graham'south Law:

r₁ / r_ii = SQRT(MM_ii / MM₁)

ii) Enter appropriate values into the above equation:

ane / 2.92 = SQRT(ten / 17.0307)

The sub-ones are for the ammonia and the sub-twos are the unknown gas.

3) Square both sides (remember example #6!):

0.1172828 = ten / 17.0307

x = 2.00 g/mol

Although non asked for, the gas is hydrogen (H₂).

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Case #10: If a molecule of C_twoH₆ diffuses a distance of 0.978 grand from a point source, calculate the distance (thou) that a molecule of CH₄ would diffuse under the same conditions for the same period of time.

Solution:

We can plow the 0.978 m into a rate by assuming some measure out of fourth dimension equal to one, say 1 memtosecond. So, the C₂H_vi diffuses at the rate of 0.978 yard / 1 memtosecond. When nosotros practise the adding, the other rate will be our respond since it will be the distance traveled in 1 memtosecond also.

Past the manner, the unit of 1 memtosecond is completely simulated. Notwithstanding, what nosotros do know is that C_iiH₆ does diffuse 0.978 m in some span of fourth dimension and it is that bridge of time that we are calling 1 memtosecond. This is being done to plough a distance (0.978 m) into a rate (0.978 m / memtosecond).

r_i / r₂ = $\sqrt{\mathrm{MM₂\; /\; MM_ane}}$

x / 0.978 = $\sqrt{\mathrm{30.07\; /\; 16.043}}$

x = 1.83 m

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Instance #eleven: A sample of oxygen gas (O₂) effuses into a vacuum 1 times faster than an unknown gas. O_ii has a molecular weight of about 32.00 g mol¯^ane. What is the molecular weight of this unknown gas (in 1000 mol¯^ane)?

Discussion: The student that posted this question to an "answers" website wrote: When I tried calculating this problem using the formula, I got the same reply of 32.00 g/mol, seeing as the [effusion] is merely "1 time faster". Is this correct?

Solution (not done by the ChemTeam):

I would read that as unknown gas effuses at a rate = ten, while O₂ effuses at a rate of "x" faster than unknown gas.

Therefore, . . . rate O_ii = x + x = 2x

Use Graham'due south Law

(charge per unit unknown / charge per unit O₂) = SQRT[32 / Z)]

(10 / 2x) = SQRT[32 / Z)]

(1/2)² = 32 / Z

Z = 4 * 32 = 128 g/mol

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Case #12: Argon effuses from a container at a rate of 0.0260 L/s. How long will it have for 3.00 50 of I_ii to effuse from the aforementioned container under identical conditions?

Solution:

1) State Graham's Police:

r₁/r₂ = SQRT[MM₂ / MM_i]

2) State the molar masses:

MM of Ar = 39.948 g/mol

MM of I₂ = 253.8 1000/mol

3) Use r₂ and MM₂ for argon.

r₁ / 0.0260 = SQRT[39.948 / 253.8]

Notation that the g/mol unit on each MM cancels.

r_ane / 0.0260 = 0.39674

r₁ = 0.01031524 L/s

4) The time needed for 3.00 L to effuse:

3.00 Fifty / 0.01031524 L/south = 291 south (to three sig figs)

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Example #thirteen: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of oxygen at the same temperature and pressure level, effuses through this apparatus in 8.fifty minutes.

Solution #1:

1) Write Graham'due south Police:

$\text{r}{}_1^{\mathrm{two}}$ / $\text{r}{}_{\mathrm{two}}^2$ = MM_two / MM_one

2) Identify values to be used:

r₁ (oxygen) ---> 1 / 8.fifty min
r₂ (unknown) ---> one / 6.60 min

MM_ane = 32.00 1000/mol
MM₂ = ten

Note how 'i' is used as a placeholder for the 'given volume' specified in the problem statement. Sometimes, I volition also only assume the given value was one Fifty. Information technology doesn't matter what the actual book is, because the only requirement is that the two volumes be equal.

three) Insert values and solve:

(1 / eight.50)² / (1 / vi.threescore)² = x / 32.00

6.threescore² / 8.50² = x / 32.00

43.56 / 72.25 = x / 32.00

10 = xix.293 chiliad/mol

four) Given that molar volume is 22.414 Fifty at STP, calculate the density of the gas:

xix.293 g/mol / 22.414 L/mol = 0.862 g/50

Solution #2:

This is the solution given by the person (not the ChemTeam) who answered the question on an "answers" website:

According to Graham's law of effusion:

Effusion rate ∝ 1/$\sqrt{\mathrm{(Tooth\; mass)}}$ . . . . . . . [i]

When going through an apparatus:

Effusion charge per unit ∝ 1/$\sqrt{\mathrm{(Time\; taken)}}$ . . . . . . . [2]

Compare [1] and [ii]:

$\sqrt{\mathrm{(Molar\; mass)}}$ ∝ (Time taken), i.eastward.

$\sqrt{\mathrm{(Tooth\; mass\; of\; the\; gas)}}$ / $\sqrt{\mathrm{(Tooth\; mass\; of\; O₂)}}$ = (Time taken for the gas) / (Time taken for O₂)

$\sqrt{\mathrm{(Tooth\; mass\; of\; the\; gas)}}$ / $\sqrt{\mathrm{(32.0\; g/mol)}}$ = (6.lx min) / (8.50 min)

Tooth mass of the gas = 32.0 x (6.lx/8.50)^two g/mol

Molar mass of the gas = 19.3 g/mol

Molar volume of a gas at STP = 22.4 L/mol

Density of the gas at STP = (19.3 thou/mol) / (22.4 L/mol) = 0.862 g/L

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Bonus Instance #ane: The charge per unit of effusion of an unknown gas at 480 G is i.6 times the rate of effusion of SO₂ gas at 300 K. Calculate the molecular weight of the unknown gas.

Solution:

For a give-and-take about the equation that will exist used to solve this trouble, please get here.

1) We will use the 2d equation, but without the V_i and Five₂.

r_one / r₂ = $\sqrt{\mathrm{(T_iM_two)\; /\; (T₂Grand_one)}}$

2) We prepare the charge per unit of effusion for SO₂ to be equal to i. That means the charge per unit of effusion for the unknown gas is one.half-dozen. Let us utilise r₂ for the And then₂:

1.half-dozen / 1 = $\sqrt{\mathrm{(480\; <b>·</b>\; 64.063)\; /\; (300\; <b>·</b>\; x)}}$

iii) Square both sides:

2.56 = (480 · 64.063) / (300 · 10)

4) Multiply and cross-multiply:

30750.24 = 768x

x = 40 g/mol

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Bonus Example #two: Heavy h2o, D_twoO (tooth mass = twenty.0276 g mol¯¹), can be separated from ordinary h2o, H₂O (molar mass = 18.0152 1000 mol¯^one), as a outcome of the difference in the relative rates of diffusion of the molecules in the gas stage. Calculate the relative rates of improvidence for H₂O when compared to D₂O.

Solution:

1) Write Graham's Constabulary:

$\text{r}{}_1^2$ / $\text{r}{}_{\mathrm{ii}}^2$ = MM₂ / MM_i

Non the usual ChemTeam writing of Graham'due south Constabulary, simply it nonetheless works.

2) The D₂O is slower, so I'm going to assign it to r_two and give it a value of 1. That ways that the lighter H₂O rate will be in the numerator and exist a value greater than one.

10^two / aneⁱⁱ = 20.0276 / eighteen.0152

x^two = i.11170567

ten = one.05437454

3) H_twoO diffuses at a charge per unit one.05 times that of D₂O.

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Boyle'due south Law	No Name Constabulary	Graham'southward Police force Probs 1-x
Charles' Law	Combined Gas Law	Graham's Law Probs eleven-25
Gay-Lussac's Law	Ideal Gas Law	All Graham's Law examples and problems
Avogadro'southward Law	Dalton'south Law	Return to KMT & Gas Laws Card
Diver'south Law

A Sample Of N2 Effuses In 220 S. How Long Will The Same Size Sample Of Cl2 Take To Effuse?,

Source: https://www.chemteam.info/GasLaw/Gas-Graham.html

Posted by: closethruseell.blogspot.com

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