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R"'~ ... ... "f " '~ . "': . ,,? determine 1. I(a) .. ~ ~ ~ ... ; .. ! .. ;.... : .. i . . i ... . : .. j ••. ~ determine 1. I(b) M = RT (~) p (0. 0820615 dm3 atm mol-I okay- 1) x (273. 15 ok) x (1. 42755 g dm- three atm- l ) p=o = 131. 9987 g mol-II the worth acquired for R deviates from the approved price by way of zero. 005 percentage. the mistake effects from the truth that simply 3 information issues can be found and linear extrapolation was once hired. The molar mass, even though, concurs precisely with the approved price, most likely due to compensating plotting error. E1. 8(b) The mass density p is said to the molar quantity Vrn by way of M p the place M is the molar mass. placing this relation into the ideal gasoline legislations yields pVrn = RT so pM =RT p THE houses OF GASES 7 Rearranging this consequence supplies an expression for M; after we understand the molar mass, we will divide by means of the molar mass of phosphorus atoms to figure out the variety of atoms in line with fuel molecule + (8. 314 Pa m3 mol-I) x [(100 273) ok] x (0. 6388kgm- three ) ----------------------~----------~---- RTp M--P - 1. 60xl()4Pa = zero. 124 kg mol- I = 124 g mol- I The variety of atoms consistent with molecule is 124g mol - I ----'------,-I = four. 00 31. 0g molsuggesting a formulation of ~ E1. 9(b) Use the proper fuel equation to compute the volume; then convert to mass. PV pV = nRT so n = RT we want the partial strain of water, that is fifty three percentage of the equilibrium vapor strain on the given temperature and traditional strain. p = (0. fifty three) x (2. sixty nine x 10 three Pa) = 1. forty three x 103 Pa so n = (1. forty three x 103 Pa) x (250 m3 ) 2 = 1. forty five x 10 mol (8. 3145 J ok- I mol-I) x (23 + 273) ok or m = (l. forty five x 102 mol) x (l8. 0g mol-I) = 2. sixty one x 103 g = 12. sixty one kg 1 E1. 10(b) (a) the amount occupied via every one gasoline is identical, when you consider that every one thoroughly fills the box. hence fixing for V we have now (assuming an ideal gasoline) V nJRT zero. 225 g =- nNe = --------=-----,I 20. 18 g mol- PJ = 1. 115x 10- 2 mol, V= (!. l15 x 10- 2 PNe= eight. 87kPa, T=300K mol) x (8. 314 kPa okay- I mol-I) x three hundred okay) three =3 . 137dm eight. 87 kPa dm three =13 . 14dm3 1 (b) the whole strain is set from the complete volume of gasoline, n = nCH4 nCH4 = zero. 320 g -2 I = 1. 995 x 10 mol sixteen. 04 g mol- n = (1. 995 + zero. 438 + 1. one hundred fifteen) nAr = + nAr + nNe. zero. one hundred seventy five g = four. 38 x 1O- three mol 39. ninety five g mol- I x 1O- 2 mol = three. 548 x 1O- 2 mol p = nRT [1. eight] = (3. 548 x 10- mol) x (8. 314 d_m three kPa ok- I mol-I) x (300 ok) 2 three. 137dm three V = 128. 2 kPa 1 8 E1. 11(b) STUDENT'S suggestions handbook this can be just like workout l. ll(a) with the exception that the density is first calculated. RT M = p - [Exercise 1. 8(a)] p 33. 5mg =0. 1340gdm- three , 250cm p= ---3 M= E1. 12(b) p= 152 Torr, T=298K (0. 1340gdm- three ) x (62. 36dm3 TorrK- 1 mol-I) x (298K) I -II = sixteen. 14gmol 152 Torr This workout is identical to workout 1. 12(a) in that it makes use of the definition of absolute 0 as that temperature at which the quantity of a pattern of fuel might develop into 0 if the substance remained a gasoline at low temperatures. the answer makes use of the experimental undeniable fact that the quantity is a linear functionality of the Celsius temperature.