Pj Problems - Overview
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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)
Dalton's Law: Pt = Σi = 1-n Pi = P1 + P2 ... Pn.
Where Pt = total pressure of gas mixture; Pi = partial pressure of gasi in mixture = pressure of gasi (i.e. independent of other gas pressures in the mixture)
Mole Ratio = ni/nt = Pi/Pt.
Where ni = number of moles of gasi in the gas mixture; nt = total nuber of moles in the gas mixture
(a) Consider the following gas mixing apparatus:
(i) What is the partial pressure of N2 after the stopcock between the two containers is opened and the gases mix?
(ii) What is the partial pressure of O2 in the mixture?
(b) A mixture containing 0.538 mol of He(g), 0.315 mol of Ne(g), and 0.103 mol Ar(g) is confined in a 7 L vessel at 25oC.
(i) Calculate the partial pressures of each of the gases in the mixture
(ii) Calculate the total pressure of the mixture.
(c) A piece of solid carbon dioxide with a mass of 5.50 g is placed in a 10 L vessel that already contain air at 705 torr and 24oC. After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at 24oC?
(d) A mixture of gases contains 0.75 mol N2, 0.30 mol O2, and 0.15 mol of CO2. If the total pressure of the mixture is 1.56 atm, what is the partial pressure of each component?
(e) At an underwater depth of 250 ft, the pressure is 8.38 atm. What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm.
(f) A quantity of N2 gas originally held at 4.75 atm pressure in a 1 L container at 26oC is transferred to a 10 L container at 20oC. A quantity of O2 gas originally at 5.25 atm and 26oC in a 5 L container is transferred to this same container. What is the total pressure in the new container?
S7P3A32 (force - push).
Pj Problem of Interest is of type force (push).
(a) n = PV/RT calculates the initial moles of N2 and O2 in the apparatus:
So, nN2 = 1x2/(0.0821x298) = 2/24.465 = 0.082 mol of N2
So, nO2 = 2x3/(0.0821x298) = 6/24.465 = 0.245 mol of O2
P = nRT/V (note new volume of gases = 5 L)
So, PN2 = (0.082x0.0821x298)/5 = 0.40 atm
So, PO2 = (0.245x0.0821x298)/5 = 1.20 atm
So, Pt = PN2 + PO2 = 0.40 + 1.20 = 1.60 atm.
(b) P = nRT/V
So, PHe(g) = (0.538x0.0821x298)/7 = 1.88 atm
So, PNe(g) = (0.315x0.0821x298)/7 = 1.10 atm
So, PAr(g) = (0.103x0.0821x298)/7 = 0.360 atm
So, Pt = 1.88 + 1.10 + 0.360 = 3.34 atm.
(c) P = nRT/V
Molar mass of CO2 = 44 g
So, mol of 5.5 g = 5.5/44 = 0.125 mol
So, PCO2 = 0.125x0.0821x297)/10 = 0.305 atm
Pair = 705/760 atm = 0.928 atm
So, Pt = 0.305 + 0.928 = 1.233 atm.
(d) Pi = (ni/nt)Pt.
nt = 0.75 + 0.30 + 0.15 = 1.20 mol
Pt = 1.56 atm
So, PN2 = (0.75/1.20)1.56 = 0.975 atm
So, PO2 = (0.30/1.20)1.56 = 0.39 atm
So, PCO2 = (0.15/1.20)1.56 = 0.195 atm.
(e) Pi = (ni/nt)Pt
Mole percent = (ni/nt)100
PO2 = 0.21 atm
Pt = 8.38 atm
Pi = (ni/nt)Pt
So,(ni/nt)100 = (0.21/8.38)100 = 2.5%.
(f) n = PV/RT
So, original nN2 = (1x4.75)/(0.0821x299) = 4.75/24.548 = 0.1935 mol
So, original nO2 = (5.25x5)/(0.0821x299) = 26.25/24.548 = 1.069 mols
After transfer to 10 L container:
PN2 = nRT/V = (0.1935x0.0821x293)/10 = 0.466 atm
PO2 = nRT/V = (1.069x0.0821x293)/10 = 2.572 atm
So, Pt = 0.466 + 2.572 = 3.038 atm.
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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