Tutorial 1 answers 2020 A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate PDF

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CMY282 – PHYSICAL CHEMISTRYTutorial 1: Gas LawsQUESTION 1A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.Solution. We are given the masses of the three...

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CMY282 – PHYSICAL CHEMISTRY Tutorial 1: Gas Laws QUESTION 1 A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.

Solution.

We are given the masses of the three gasses and 𝑝𝑁𝑒 = 8.87 kPa at 300 K. (a) The volume occupied by each gas is the same as they fill the container completely. The number of moles of Ne gas is 𝑛𝑁𝑒 =

225 × 10−3 g = 0.01115 mol 20.18 g ⋅ mol−1

The volume of the container can be obtained through using the ideal gas law, but with the pressure and number of moles specific for neon gas 𝑉=

𝑛𝑁𝑒 𝑅𝑇 (0.01115 mol)(8.314 dm3 ⋅ kPa ⋅ K −1 ⋅ mol−1 )(300 K) = 3.14 dm3 = 3.14 L = 8.87 kPa 𝑝𝑁𝑒

(b) The total pressure is given by 𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝𝐶𝐻4 + 𝑝𝐴𝑟 + 𝑝𝑁𝑒 =

𝑛𝑡𝑜𝑡𝑎𝑙 𝑅𝑇 𝑉

The total number of moles is just the sum of all the moles of the three gasses. So, we need 𝑛𝐶𝐻4 = 𝑛𝐴𝑟 =

320 × 10−3 g = 0.01995 mol 16.04 g ⋅ mol−1

175 × 10−3 g = 0.00438 mol 39.95 g ⋅ mol−1

and 𝑛𝑁𝑒 = 0.01115 mol from (a), so the total number of moles is

𝑛𝑡𝑜𝑡𝑎𝑙 = (0.01115 + 0.01995 + 0.00438) = 0.03548 mol

From Dalton’s law we know that

𝑝𝑁𝑒 = ( so then 𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝𝑁𝑒 ( Or you could use the ideal gas law 𝑝𝑡𝑜𝑡𝑎𝑙 =

𝑛𝑡𝑜𝑡𝑎𝑙 𝑅𝑇 𝑉

=

𝑛𝑁𝑒 )𝑝 𝑛𝑡𝑜𝑡𝑎𝑙 𝑡𝑜𝑡𝑎𝑙

𝑛𝑡𝑜𝑡𝑎𝑙 0.03548 mol ) = 8.87 kPa ( ) = 28.2 kPa 𝑛𝑁𝑒 0.01115 mol

(0.03548 mol)(8.3145 dm3 ⋅ kPa ⋅ L−1 ⋅ mol−1 )(300 K) = 28.2 kPa 3.14 dm3

1

QUESTION 2 At the stratospheric altitude of about 30 km the average pressure is 50 Torr while the average temperature is 230 K. Assuming that the atmosphere at this altitude consists mainly of N 2 molecules with a collision diameter of 395 pm, calculate (a) the mean speed of the molecules, (b) the mean free path, (c) the collision frequency in the stratospheric gas.

Solution. (a) The mean speed can be calculated from 𝜈𝑚𝑒𝑎𝑛

1

8(8.314 J ⋅ K −1 ⋅ mol−1 )(230 K) 2 8𝑅𝑇 2 =( ) =( ) = 417 m ⋅ s −1 𝜋𝑀 𝜋(28.02 × 10−3 kg ⋅ mol−1 ) 1

(b) The mean free path can be calculated from 𝜆=

𝑘𝑇 𝑘𝑇 = = 𝜎𝑝 𝜋𝑑 2 𝑝

𝜋(395 ×

(1.381 × 10−23 J K −1 ) ⋅ (230 K)

10−12 m)2

1 atm 101325 Pa (50 Torr ( )( )) 760 Torr 1 atm

= 972 nm

(Note: 1 J = 1 kg m2 s-2 and 1 Pa = 1 kg m−1 s-1 ) (c)

Here we employ the equation 𝜆=

𝜈𝑟𝑒𝑙 𝑧

⇒

𝑧=

√2 (417 m ⋅ s−1 ) 𝜈𝑟𝑒𝑙 √2 𝜈𝑚𝑒𝑎𝑛 = 6.06 × 108 s−1 = = 972 × 10−9 m 𝜆 𝜆

2

QUESTION 3 Use the Maxwell distribution of speeds to estimate the fraction of N 2 molecules at 500 K that have speeds in the range 290 to 300 m s-1.

Solution. Start by assuming that the range of speed is narrow. Armed with this assumption, we may approximate the fraction of molecules with FractionN2 = 𝑓(𝑣) × 𝛥𝑣

Though 𝑣 varies between 290 and 300 m⋅s–1, the variation of 𝑓(𝑣) is assumed to be small over this temperature range, and the speed value at the mid-point (centre) of the speed range is used in the calculation. This mid-point is given by 𝑣=

300 + 290 = 295 m ⋅ s−1 2

The Maxwell distribution is given by the equation

3

M 2 2 −M𝑣2 𝑓(𝑣) = 4π ( ) 𝑣 e 2RT 2πRT To simplify the calculation, we may calculate some terms separately. First,

28.02 × 103 kg ⋅ mol−1 M = = 3.37 × 10−6 m−2 ⋅ s2 2RT 2(8.314 J ⋅ K −1 ⋅ mol−1 )(500 K)

Next, 𝑓(295 m ⋅ s

−1 )

= =

(Note: 1 J = 1 kg m2 s-2)

3

3.37 × 10−6 m−2 ⋅ s2 2 −6 4π ( ) (295 m ⋅ s −1 )2 e−(3.37×10 π

m −2⋅ s2)(295 m⋅s−1)

2

9.06 × 10−4 m−1 ⋅ s

So, then the fraction of molecules in the range 290 to 300 m⋅s–1 is given by 𝑓(𝑣) × 𝛥𝑣

= = = =

(9.06 × 10−4 m−1 ⋅ s ) × (300 m ⋅ s−1 − 290 m ⋅ s−1 ) (9.06 × 10−4 m−1 ⋅ s ) × (10 m ⋅ s−1 ) 9.06 × 10−3 0.91%

(This is a fraction, therefore × 100 to get %)

Note that the small value obtained (0.91%) for the percentage of N2 molecules traveling with speeds between 290 to 300 m s-1 suggests that the approximation, 𝑓(𝑣) is constant over the range, is valid. An alternative test is to look at the values of 𝑓(300) and 𝑓(290), and see by how much they differ. If they do not differ too much, the function is constant over that range. In general, we will assume that if we obtain a final answer less than 5%, our simplified method using the assumption 𝑓(𝑣) × 𝛥𝑣 to calculate the fraction of molecules, is valid.

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QUESTION 4 To what temperature must He atoms be cooled so that they have the same r.m.s. speed as O2 at 25°C?

Solution. The root mean square speed is given by the equation 𝜐𝑟𝑚𝑠 = (

1

3𝑅𝑇 3𝑅𝑇 2 ) = √ 𝑀 𝑀

So, we need the temperature (THe) at which he atoms will have the same value for vrms as it does for O2 at 298.15 K. Then, to determine THe , we must compare the root mean square speed of the He atoms with that of the oxygen molecules as follows 𝜐𝑟𝑚𝑠 (O2 )

∴√

3𝑅𝑇𝑂2 𝑀𝑂2

∴

TO2 MO2

∴ THe

=

𝜐𝑟𝑚𝑠 (He)

=

√

=

3𝑅𝑇𝐻𝑒 𝑀𝐻𝑒

THe MHe

=

MHe TO2

=

(298.15 K)(4.00 g ⋅ mol−1 )

=

MO2

31.998 g ⋅ mol−1

37.7 K

(or − 261o C)

4

(Solving for THe )

QUESTION 5 A gas at 350 K and 12 atm has a molar volume 12 percent larger than that calculated from the perfect gas law. Calculate (a)

the compression factor under these conditions, and

(b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces?

Solution.

(a) The molar volume of the real gas (Vm) is 12% larger than that of a perfect gas (𝑉𝑚∘) under the same conditions. Hence, from the definition of the compression factor we can see 𝑍=

𝑉𝑚 (𝑉𝑚∘ + 0.12𝑉𝑚∘ ) = = 1.12 𝑉𝑚∘ 𝑉𝑚∘

(b) To calculate the molar volume of the gas (Vm), we can use the compression factor in combination with the molar volume for an ideal gas. So, we start with 𝑉𝑚∘ =

𝑅𝑇 (0.08206 𝐿 𝑎𝑡𝑚 𝐾 −1 𝑚𝑜𝑙 −1 )(350 𝐾) = 2.39 𝐿 ⋅ 𝑚𝑜𝑙 −1 = 12 𝑎𝑡𝑚 𝑝

Then, we can obtain the molar volume of the gas from

𝑉𝑚 = 𝑍 ⋅ 𝑉𝑚0 = (1.12)(2.39 𝐿 ⋅ 𝑚𝑜𝑙 −1 ) = 2.68 𝐿 ⋅ 𝑚𝑜𝑙 −1 ≈ 2.7 𝐿 ⋅ 𝑚𝑜𝑙 −1

Under these conditions, Z > 1, hence repulsive forces dominate. The repulsive forces in the real gas causes the molar volume of the real gas to be larger than that of the perfect gas at the same temperature and pressure (𝑉𝑚 > 𝑉𝑚∘).

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QUESTION 6 At 300 K, the virial coefficient (B) of N2 and CH4 are –4.2 cm3 ⋅mol–1 and –15 cm3 mol–1 respectively. Which gas behaves more ideally at this temperature?

Solution.

So, to clarify B(N2) = –4.2 cm3 ⋅mol–1 and B(CH4) = –15 cm3 mol–1. Also, remember that 𝑍 = 1 for an ideal gas. Deviation from

Z = 1 is used as a measure of departure from perfect gas behavior. We can therefore use the value of Z to compare gasses to the ideal gas. For the virial equation of state: 𝑍=

𝑝𝑉𝑚 𝑅𝑇

= 1+

𝐵

𝑉𝑚

+⋯

In order to compare the expressions for Z for these gases, we need to assume that 𝑉𝑚 (N2 ) ≈ 𝑉𝑚 (CH4 ). Note that this may

introduce a small error, as the molar volumes of these gases do differ slightly. At 298 K and 1 atm, the molar volume of N 2 is

22.4 L mol-1 while that of CH4 is 24.4 L mol-1. Unfortunately, this information was not provided for the conditions listed in this question, therefore we will assume that the contribution from the small difference in molar volume of these gases will be minimal. Since the second virial coefficient (B) is negative for both gases, Z 1, repulsive forces in the real gas dominate causing 𝑉m to be more than 𝑉m∘.

So, due to the relationship between Z and B, when B is positive, Z > 1 and repulsive forces will dominate (i.e. at 373 K for this problem). B is negative, 𝑍 < 1 and attractive forces will dominate (i.e. at 273 K for this problem).

7

QUESTION 8 Suppose that 10.0 mol C2H6(g) is confined to 4.860 dm3 at 27°C. Predict the pressure exerted by the ethane and calculate the compression factor (a) assuming ethane to behave like a perfect gas, and (b) using the van der Waals equation of state. Which intermolecular forces will dominate under these conditions? For ethane, a = 5.489 dm6 atm mol–2, b = 0.06380 dm3 mol–1.

Solution. (a) For a perfect gas, Z = 1. Next, we can write the perfect gas law in terms of pressure and substitute the given values as follows 𝑝=

𝑛𝑅𝑇

=

𝑉

(10.0 mol)(0.08206 dm3 ⋅ atm ⋅ K−1 ⋅ mol−1 )(300 K) 4.860 dm3

= 50.7 atm

(b) The pressure of a van der Waals gas can be written as the subject of the van der Waals Equation as follows (1 L = 1 dm 3) 𝑝

=

= = =

𝑛𝑅𝑇 𝑛2 −𝑎 2 𝑉 𝑉 − 𝑛𝑏

(10.0 mol)(0.08206 dm3 ⋅atm⋅K−1⋅mol−1)(300 K) 4.860 L − (10.0 mol)(0.06380 dm3 ⋅mol–1)

58.49 𝑎𝑡𝑚 − 23.32 𝑎𝑡𝑚 35.2 atm

Then the compression factor is 𝑍=

𝑝𝑉𝑚 𝑅𝑇

− 5.489 dm6 ⋅ atm ⋅ mol–2 (

= =

(35.2 atm)(0.486 L ⋅ mol−1 )

(0.08206 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(300 K)

0.695

Hence, since Z < 1 attractive forces dominate in the real gas as compared to if it were a perfect gas.

8

2 10.0 mol ) 4.860 dm3

QUESTION 9 Cylinders of compressed gas are typically filled to a pressure of 200 bar. For ethylene, C 2H4, what would be the molar volume at this pressure and 30°C based on (a) the perfect gas equation, (b) the van der Waals equation. For ethylene, a = 4.552 dm6 atm mol–2, b = 0.0582 dm3 mol–1.

Solution. (a) We can calculate the molar volume of an ideal gas as follows 𝑉𝑚0 =

𝑅𝑇 (0.08314 dm3 ∙ bar ∙ K −1 ⋅ mol−1 )(303.15 K) = 1.26 × 10−4 m3 ⋅ mol−1 = 0.126 dm3 ⋅ mol−1 = 200 bar 𝑝

Note: you should obtain the same answer when converting the pressure to atmosphere and use R = 0.08206 L atm K−1 mol−1

(b) Using the iteration method, we can calculate the molar volume of a van der Waals gas as follows. We start with the expression for the pressure of the gas and rewrite it in terms of the molar volume as follows 𝑝=

𝑛𝑅𝑇 𝑛2 −𝑎 2 𝑉 𝑉 − 𝑛𝑏

⇒

𝑉𝑚 =

𝑅𝑇 𝑎 + 𝑏 𝑝+ 2 𝑉𝑚

For the first iteration, we use the molar volume of a perfect gas under the same conditions to calculate a new value for Vm. It makes sense to convert the unit of pressure to atm and use the corresponding value of R, since the units of a is dm6 atm mol–2 𝑉m,1

=

=

(0.08206 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(303.15 K) + 0.0582 dm3 ⋅ mol–1 4.55 dm6 ⋅ atm ⋅ mol−2 197.4 atm + (0.126 dm3 ⋅ mol−1 )2 0.110 dm3 ⋅ mol−1

Then, we do the same calculation again, but using 𝑉m,1 as our estimation 𝑉m,2

=

=

(0.08206 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(303.15 K) + 0.0582 dm3 ⋅ mol–1 4.55 dm6 ⋅ atm ⋅ mol−2 197.4 atm + (0.110 dm3 ⋅ mol−1 )2 0.102 dm3 ⋅ mol−1

and finally, the third iteration using 𝑉m,2 𝑉m,3

=

=

(0.08206 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(303.15 K) + 0.0582 dm3 ⋅ mol–1 4.55 dm6 ⋅ atm ⋅ mol−2 197.4 atm + (0.102 dm3 ⋅ mol−1 )2 0.0974 dm3 ⋅ mol−1

Note: for the purpose of answering a question in a test or exam, you will be required to perform only 3 iterations. However, if you require an accurate answer you will continue repeating the iterations until the answer obtained for the molar volume converges (i.e. remains the same)

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