Chapter 19

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4.A quantity of ideal gas at 7°C and 51 kPa occupies a volume of 2.6 m³. (a) How many moles of the gas are present? (b) If the pressure is now raised to 210 kPa and the temperature is raised to 27.0°C, how much volume does the gas occupy? Assume no leaks.

21.The lowest possible temperature in outer space is 2.75 K. What is the rms speed of hydrogen molecules at this temperature? (The molar mass of molecular hydrogen is 2.02 g/mol )

26. What is the average translational kinetic energy of nitrogen molecules at 2140 K?

30. The mean free path of nitrogen molecules at 24°C and 1.52 atm is 0.251 x 10^-5 cm. At this temperature and pressure there are 3.76 x 10¹⁹ molecules/cm³. What is the molecular diameter, in cm?

33.The speeds of 10 molecules are 2.0, 3.0, 4.0, . . . , 11 km/s. In km/s, what are their (a) average speed and (b) rms speed?

34.In fig. 19-25, 1.43 mole of an ideal diatomic gas can go from a to c along either the direct (diagonal) path ac or the indirect path abc. The scale of the vertical axis is set by p_ab = 6.27 kPa and p_c = 2.28 kPa, and the scale of the horizontal axis is set by V_bc = 4.67 m³ and V_a = 2.34 m³. (The molecules rotate but do not oscillate.) What is the net transfer of energy as heat for the indirect path?


ANSWERS
4. (a) With T = 283 K, we obtain

pV (100´ 103Pa)(2.50m3)
n = = = 106mol.
RT (8.31J/mol×K)(283K)

(b)  We  can  use  the  answer  to  part  (a)  with  the  new  values  of  pressure  and
temperature, and solve the ideal gas law for the new volume, or we could set up the
gas law in ratio form as:
p V T
f f f

pV T
i i i

(where ni = nf and thus cancels out), which yields a final volume of 

T 3100kPa 303K 3
VfVipi f 2.50m 0.892 m .
pfTi 300kPa 283K

21. According to kinetic theory, the rms speed is

3RT
v 
rms
M

where T is the temperature and M is the molar mass. See Eq. 19-34. According to
–3
Table  19-1,  the molar mass  of molecular  hydrogen  is  2.02  g/mol  =  2.02    10 
kg/mol, so

3 8.31J/mol K 2.7K
2
v 1.8 10 m/s.
rms 3
2.02 10 kg/mol

Note: The corresponding average speed and most probable speed are

RT 8 8.31J/mol K 2.7K
8 2
v 1.7 10 m/s
avg 3
M (2.02 10 kg/mol)
and
1
 
RT 2 8.31J/mol K 2.7K
2 2
vp 1.5 10 m/s ,
3
M 2.02 10 kg/mol
respectively.
26. The average translational kinetic energy is given by  Kavg 32 kT , where k is the
–23
Boltzmann constant (1.38   10  J/K) and T is the temperature on the Kelvin scale.
Thus

3 23 20
K (1.38 10 J/K)(1600K) = 3.31 10 J .
avg
2

30. We solve Eq. 19-25 for d:

11
d 
5 19 3
2(N/V) (0.80 10 cm) 2(2.7 10 /cm )

–
which yields d = 3.2   10 8 cm, or 0.32 nm.

v
33.  (a) The  average  speed  is  v ,  where  the  sum  is  over  the  speeds  of  the
N
particles and N is the number of particles. Thus

(2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0)km/s
v 6.5km/s.
10

2
v
(b) The rms speed is given by vrms     .  Now
N

2 2 2 2 2 2
v   [(2.0) (3.0) (4.0) (5.0) (6.0)

2 2 2 2 2 2 2 2 2
(7.0) (8.0) (9.0) (10.0) (11.0) ] km /s 505km /s

so
22
505km /s
v 7.1km/s.
rms
10




2
 


44. Two formulas (other than the first law of thermodynamics) will be of use to us. It
is straightforward to show, from Eq. 19-11, that for any process that is depicted as a
straight line on the pV diagram, the work is

pp
if
WV
straight
2

which includes, as special cases, W = p V for constant-pressure processes and W = 0
for constant-volume processes. Further, Eq. 19-44 with Eq. 19-51 gives

ff
E n RT pV 
int
22

where we have used the ideal gas law in the last step. We emphasize that, in order to
obtain work and energy in joules, pressure should be in pascals (N / m2) and volume
should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5.

(a) The internal energy change is

553 3 3 3
E E pV pV 2.0 10 Pa 4.0m 5.0 10 Pa 2.0m
int c int a c c a a
22 
3
5.0 10 J.

(b) The work done during the process represented by the diagonal path is

papc 33
W V V  =   3.5 10 Pa 2.0m 
diag ca
2

which yields Wdiag = 7.0×103 J. Consequently, the first law of thermodynamics gives

3 3 3
Q E W ( 5.0 10 7.0 10 ) J 2.0 10 J. 
diag  int diag

(c) The  fact  that  Eint only depends on  the  initial and  final  states, and not on  the
details  of  the  “path”  between  them,  means  we  can  write
3
E E E 5.0 10 J  for the indirect path, too. In this case, the work done
int int cint a
3
 
consists  of  that  done  during  the  constant  pressure  part  (the  horizontal  line  in  the
graph) plus that done during the constant volume part (the vertical line):

3 3 4
W 5.0 10 Pa 2.0m 0 1.0 10 J.
indirect

Now, the first law of thermodynamics leads to

3 4 3
Q E W ( 5.0 10 1.0 10 ) J 5.0 10 J.