Pj Problems - Overview
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Triadic Unit Mesh
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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 0oC and 1 atm. Vessel B contains SO2(g) at 20oC 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 25oC: 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.
S7P4A41 (motion - linear)
S7P1A15 (containership - mass)
S7P5A51 (change - physical)
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 nA and nB respectively
From the ideal gas equation, nA = PAVA/RTA
So, nA = nCO = (1 x V)/273R.
Similarly, nB = nSO2 = 0.5xV/(293R).
So, nA/nB = [(1 x V)/273R]/[0.5 x V/(293R)] = 2(293)/273 = 2.147
So, nA = 2.147 nB
So, vessel A contains more molecules.
(aii) Molar mass of CO = 12 + 16 = 28 g
So, mass of nA mol of CO = mA = 28nA = 28(2.147nB) =60.116nB g
Molar mass of SO2 = 32 + 32 = 64 g
So, mass of nB mol of SO2 = mB = 64nB g
So, vessel B contains more mass since mB > mA
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, rmsA = uA = [(3x8.314x273)/0.028]1/2 = (2.43 x 105)1/2 m/s
So, rmsB = uB = [(3x8.314x293)/0.064]1/2 = (1.14 x 105)1/2 m/s
So, uA/uB = 1.46
So, uA > uB.
(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
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) rmsNF3 = uNF3 =[(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, uNF3 =[(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.
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 rx
Let rate of effusion per liter of O2 be rO2.
Then, rx = 1/105 L/s
And rO2 = 1/31 L/s
So, rO2/rx = (Mx/MO2)1/2
So, (105/31)2 = Mx/MO2 = Mx/32
So, 32(3.387)2 = Mx = 32x11.47 = 367 g
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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