(j) If:: C{;r Z 12- -tx z.

d'f d (ax 2) -e-i-xZ- +J7- cLx

ol (.e-j!;(2)

o{::£

/ d) --t-x l 2-l - 2 h x.J2 0.. ;;L

2-az

a J x""dx ::__0_ ?C/h--f1 f C./Yl -f -1

>;:c -10 f x M d x - J () ;( ""d ) 3:: _-1_0 t-4 fi/l f A + 3/tl +~

/Yl of A 1 A'l -f .11

2.02 0 y: -10 f )( ~ ~ -10 - g 0-1

4 x ·1 - '1 t< 1010

10- 41

@ A'l 0 -I /Mol -> f' 0" 0 OJ 2 4 L_ 01 A'vl I fi1I<of. k

0.) T = 21 3 k ) P::. 1 (;{.f- /YVt ::.) V:: 41 12T _.~(M'IC)) 'I 0 0~2 1(.L. cd /YVt I/YY1CJ k );?;

P~0~hVl)

22.4 (L)~) T:: 2 9f t 7 P::. -1oJ./YV' :::-) V :::.A/l 12T ::!!. /J'VlCJlj y 0 0R2 1 CL. Off /fvl ( /Y'v1(J!. . k-) >( 2 q P k:

f - --------1 ((}(f ftvt )

24 4 (L)

c)T:: ~1-3 t 7 p~ 1otl/YVl

;::)\I :: /VI £T = .1 (/VVtcl) 'f O. 0 f 2 -1 (L ext MA./ /W1oj _ k J y: 3l-.3 tP 1(){}/W1

~0 _6' 2 (L)

2.43 Using the data shown in Table 2.1, calculate the pressure exerted by 2.500 moles of carbondioxide confined in a volume of 1.000 L at 450 K. Compare the pressure with that calculatedassuming ideal behavior.

For CO2, a = 3.60 atmL2 mol-2 and b = 0.0427 Lmol-1. Rearrange the van der Waals equation,

(p +~) (V - nb) = nRT to give

nRT an2

p=--V - nb V2

(2.500 mol) (0.08206 L atm K-I mol-I)(450 K)=

1.000 L - (2.500 mol) (0.0427 L mol-I)

(3.60 atm L2 mol-2)(2.500 mol)2(1.000 L)2

nRT (2.500 mol) (0.08206LatmK-1 mol-I) (450K)p = -- = ----------------- = 92.3 atm

V 1.000 L

The pressure calculated using the van der Waals equation of state is smaller than that calculatedusing the ideal gas equation. Thus, under these conditions there are net attractive forces betweenmolecules.

Th d . ed I' h" vnns Hee eSlr re atlOns Ip ISVnns He = Vnns 0 ' or --' - = 1.Write the v 's in this relationship, "Vnns 0 nns

in terms of M and T: ' ,

J3~THeVnns, He He---= =J3RT

O,~

MH (4.003gmol-1)THe = Mo~ TO, = 32.00 g mol-I (273 + 25) K = 37.3 K

2.63 The speeds of 12 particles (in em s-I) are 0.5, 1.5, 1.8, 1.8, 1.8, 1.8,2.0,2.5,2.5,3.0,3.5,and 4.0. Find (a) the average speed, (b) the root-mean-square speed, and (c) the most probablespeed of these particles. Explain your results.

12LC;

c= ;=1N

(0.5 + 1.5+ 1.8+ 1.8+ 1.8+ 1.8+ 2.0 + 2.5 + 2.5 + 3.0 + 3.5 + 4.0) cm S-l

•. 12

12

LCfc2 = ;=1

N

(0.52 + 1.52 + 1.82 + 1.82 + 1.82 + 1.82 + 2.ol + 2.52 + 2.52 + 3.ol + 3.52 + 4.02) cm2 s-z

12

Cnns = j7i = 2.4 cm s-l

(c) crop = 1.8 cm S-l, as this is the speed that appears most frequently.

As expected, Cnns > c. However, because 12 particles do not constitute a macroscopic system,crop can be greater or smalJer than Cnns or c.

@ 2.5 Starting with the ideal-gas equation, show how you can calculate the molar mass of a gas froma knowledge of its density.

mPV =nRT = -RT

M

M= m RT = pRTV P P

2.17 Nitrogen dioxide (N02) cannot be obtained in a pure form in the gas phase because it exists asa mixture of N02 and NZ04• At 25°C and 0.98 atm, the density of this gas mixture is 2.7 g L-1.

What is the partial pressure of each gas?

Calculate the average molar mass, Mmix' of the mixture using the relation between M, p, T, andP derived in Problem 2.5. This average molar mass is related to the mole fraction of N02, xNO"and the mole fraction ofN204, xN,O.' The mole fractions can then be used to calculate the partialpressures of the gases.

PmixRT (2.7 g L-I) (0.08206 L atm K-1 mol-I) (273 + 25) K -I:Mnux = --- = ----------------- = 67.4 g mol

Pmix 0.98 atm

xNo,MNO, + xN,o.MN,o. = 1v1mix = 67.4 g mol-I

The sum of all mole fractions is unity, that is, xNO, + xN,O. = 1,or xN,O. = 1 - xNO,' Using thisrelation, the above equation becomes

xNo,MNO, + (I - XNO,) MN,o. = 67.4 g mol-I

xNO, (46.01 g mol-I) + (I - XNO,) (92.02 g mol-I) = 67.4 g mol-I

46.01xNo, + 92.02 - 92.02xNO, = 67.4

46.0lxNO, = 24.6

XNO, =0.535

PNO, = XNO,Pmix = (0.535) (0.98 atm) = 0.52 atm

PN,O. = xN,O. Pmix = (0.465) (0.98 atm) = 0.46 atm