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Calculation Of Average Kinetic Energy, Effusion Rate And Molar Mass Of A Gas
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Calculation Of Average Kinetic Energy, Effusion Rate And Molar Mass Of A Gas

Given that volume = V liters; pressure = P atmosphere; temperature = T Kelvin; Numuber of Moles = n.

Boyle's Law: V α 1/P (constant n, T)

Charle's Law: V α T (constant n, P)

Avogadro's Law: V α n (constant P, T)

Ideal Gas Equation: V α nT/P ; V = nRT/p (R is proportionality constant = 0.0821 L-atm/mol-K

Graham's Law Of Effusion: the effusion rate of a gas is inversely proportional to the square root of its molar mass.

(a) Vessel A contains CO₂(g) at 0^oC and 1 atm. Vessel B contains SO₂(g) at 20^oC and 0.5 atm. The two vessels have the same volume:
(i) Which vessel contains more molecules?
(ii) Which vessel contains more mass?
(iii) In which vessel is the average kinetic energy higher?
(iv) In which vessel is the rms speed of the molecules higher?
(bi) Place the following gases in order of increasing average molecular speed at 25^oC: Ne, HBr, SO₂, NF₃, CO.
(ii) Calculate the rms speed of NF₃ molecules at 25⁰C.
(c) Hydrogen has two naturally occurring isotopes.¹H and ²H. Chlorine also has two naturally occurring isotopes, ³⁵Cl and ³⁷Cl. Thus hydrogen chloride consists of four distinct types of molecules: ¹H³⁵Cl, ¹H³⁷Cl, ²H³⁵Cl, and ²H³⁷Cl. Place these four molecules in order of increasing rate of effusion.
(d) A gas of unknown molecular mass was allowed to effuse through a small opening under constant pressure conditions. It required 105 seconds for 1 L of the gas to effuse. Under identical experimental conditions it required 31 seconds for 1 L O₂ gas to effuse. Calculate the molar mass of the unknown gas.

The strings: S₇P₄A₄₁ (motion - linear)
S₇P₁A₁₅ (containership - mass)
S₇P₅A₅₁ (change - physical)

The math:
There are three Pj Problems of interest: kinectic energy which is energy due to motion(S₇P₄A₄₁), root mean speed which is due to change in speed (S₇P₅A₅₁), and mass of gas which is a containership problem (S₇P₁A₁₅).

(ai) Let the number of moles of gas in vessel A and vessel B be n_A and n_B respectively
From the ideal gas equation, n_A = P_AV_A/RT_A
So, n_A = n_CO = (1 x V)/273R.
Similarly, n_B = n_SO2 = 0.5xV/(293R).
So, n_A/n_B = [(1 x V)/273R]/[0.5 x V/(293R)] = 2(293)/273 = 2.147
So, n_A = 2.147 n_B
So, vessel A contains more molecules.

(aii) Molar mass of CO = 12 + 16 = 28 g
So, mass of n_A mol of CO = m_A = 28n_A = 28(2.147n_B) =60.116n_B g
Molar mass of SO₂ = 32 + 32 = 64 g
So, mass of n_B mol of SO₂ = m_B = 64n_B g
So, vessel B contains more mass since m_B > m_A
This solution can also be arrived at by comparing density of CO (1.25 g/L) and density of SO₂ (1.33 g/L).

(aiii) Average kinetic energy = 1/2(mu²)
Where u is the root mean speed = (3RT/M)^1/2
Where R is the constant 0.0821 L-atm/mol, T is temperature in degree Kelvin , and M is molar mass.
Average kinetic energy will be higer in vessel B which has greater mass molecules and are at higher temperature.

(aiv) Root mean speed (rms) = (3RT/M)^1/2
So, rms_A = u_A = [(3x8.314x273)/0.028]^1/2 = (2.43 x 10⁵)^1/2 m/s
So, rms_B = u_B = [(3x8.314x293)/0.064]^1/2 = (1.14 x 10⁵)^1/2 m/s
So, u_A/u_B = 1.46
So, u_A > u_B.

(bi) For an ideal gas, the average speed = 0.921(rms)
So, for gases at the same temperature, the gas with the least molar mass is the gas with the greatest molecular speed
molar masses:
Ne = 20.1797; HBR = 1 + 79.904 = 80.904; SO₂ = 32 + 32 = 64; NF₃ = 14 + 57 = 71; CO = 12 + 16 = 28.
So, in order of increasing molecular speed:
HBR < NF₃ < SO₂ < CO < Ne.

(bii) rms_NF3 = u_NF3 =[(3x8.314x298)/(71 x 10^-3)]^1/2 Where R is Joule/mol-K = 8.314 J/mol-K; and molar mass is in kg.
So, u_NF3 =[(3x8.314x298)/(71 x 10^-3)]^1/2 = (104686)^1/2 = 323.55 m/s.

(c) Molar masses:
¹H³⁵Cl = 1 + 35 = 36
¹H³⁷Cl = 1 + 37 = 38
²H³⁵Cl = 2 + 35 = 37
²H³⁷Cl = 2 + 37 = 39
By graham's law, the least mass effuse the fastest
So, the order of increasing rate of effusion is:
²H³⁷Cl < ¹H³⁷Cl < ²H³⁵Cl <sup>1H³⁵Cl.

(d)
By Graham's law rate of effusion = (k/M)^1/2
Where k is a constant, M is molar mass of gas
Let rate of effusion per liter of unknown gas be r_x
Let rate of effusion per liter of O₂ be r_O2.
Then, r_x = 1/105 L/s
And r_O2 = 1/31 L/s
So, r_O2/r_x = (M_x/M_O2)^1/2
So, (105/31)² = M_x/M_O2 = M_x/32
So, 32(3.387)² = M_x = 32x11.47 = 367 g

Math

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