19-50. (a) Using M = 32.0 g/mol from Table 19-1 and Eq. 19-3, we obtain
n
M sam
12.0 g

 0.375 mol.
M
32.0 g/mol
(b) This is a constant pressure process with a diatomic gas, so we use Eq. 19-46 and
Table 19-3. We note that a change of Kelvin temperature is numerically the same as a
change of Celsius degrees.
7 
7
Q  nC p T  n  R  T   0.375 mol    8.31 J/mol  K  100 K   1.09 103 J.
2 
 2
(c) We could compute a value of Eint from Eq. 19-45 and divide by the result from
part (b), or perform this manipulation algebraically to show the generality of this
answer (that is, many factors will be seen to cancel). We illustrate the latter approach:
Eint n  52 R  T 5
 7
  0.714.
Q
n  2 R  T 7
19-59. In the following CV  32 R is the molar specific heat at constant volume,
C p  52 R is the molar specific heat at constant pressure, T is the temperature change,
and n is the number of moles.
The process 1  2 takes place at constant volume.
(a) The heat added is
3
3
Q  nCV T  nR T  1.00 mol  8.31J/mol  K  600 K  300 K   3.74 103 J.
2
2
(b) Since the process takes place at constant volume the work W done by the gas is
zero, and the first law of thermodynamics tells us that the change in the internal
energy is
Eint  Q  3.74  103 J.
(c) The work W done by the gas is zero.
The process 2  3 is adiabatic.
(d) The heat added is zero.
(e) The change in the internal energy is
Eint  nCV T 
3
3
nR T  1.00 mol 8.31J/mol  K  455 K  600 K   1.81103 J.
2
2
(f) According to the first law of thermodynamics the work done by the gas is
W  Q  Eint  1.81103 J.
The process 3  1 takes place at constant pressure.
(g) The heat added is
Q  nC p T 
5
5
nRT  (1.00 mol) (8.31J/mol  K) (300 K  455 K)  3.22 103 J.
2
2
(h) The change in the internal energy is
Eint  nCV T 
3
3
nRT  (1.00 mol) (8.31J/mol  K) (300 K  455 K)  1.93 103 J.
2
2
(i) According to the first law of thermodynamics the work done by the gas is
W  Q  Eint  3.22 103 J  1.93 103 J  1.29 103 J.
(j) For the entire process the heat added is
Q  3.74 103 J  0  3.22 103 J  520 J.
(k) The change in the internal energy is
Eint  3.74 103 J  1.81103 J  1.93 103 J  0.
(l) The work done by the gas is
W  0  1.81103 J  1.29 103 J  520 J.
(m) We first find the initial volume. Use the ideal gas law p1V1 = nRT1 to obtain
V1 
nRT1 (1.00 mol) (8.31J / mol  K) (300 K)

 2.46  102 m3.
5
p1
(1.013  10 Pa)
(n) Since 1  2 is a constant volume process V2 = V1 = 2.46  10–2 m3. The pressure
for state 2 is
p2 
nRT2 (1.00 mol) (8.31 J / mol  K)(600 K)

 2.02 105 Pa .
2
3
V2
2.46 10 m
This is approximately equal to 2.00 atm.
(o) 3  1 is a constant pressure process. The volume for state 3 is
V3 
nRT3 (1.00 mol) (8.31J / mol  K) (455 K)

 3.73  102 m3.
p3
1.013  105 Pa
(p) The pressure for state 3 is the same as the pressure for state 1: p3 = p1 = 1.013 
105 Pa (1.00 atm)
19-60. (a) The p-V diagram is shown below:
Note that o obtain the above graph, we have chosen n = 0.37 moles for concreteness,
in which case the horizontal axis (which we note starts not at zero but at 1) is to be
interpreted in units of cubic centimeters, and the vertical axis (the absolute pressure)
is in kilopascals. However, the constant volume temp-increase process described in
the third step (see problem statement) is difficult to see in this graph since it coincides
with the pressure axis.
(b) We note that the change in internal energy is zero for an ideal gas isothermal
process, so (since the net change in the internal energy must be zero for the entire
cycle) the increase in internal energy in step 3 must equal (in magnitude) its decease
in step 1. By Eq. 19-28, we see this number must be 125 J.
(c) As implied by Eq. 19-29, this is equivalent to heat being added to the gas.
19-34. (a) From the graph we see that vp = 400 m/s. Using the fact that M = 28
g/mol = 0.028 kg/mol for nitrogen (N2 ) gas, Eq. 19-35 can then be used to
determine the absolute temperature. We obtain T =
(b) Comparing with Eq. 19-34, we conclude vrms =
1
2
Mvp2/R = 2.7×102 K.
3/2 vp = 4.9×102 m/s.
19-51. (a) Since the process is at constant pressure, energy transferred as heat to the
gas is given by Q = nCp T, where n is the number of moles in the gas, Cp is the molar
specific heat at constant pressure, and T is the increase in temperature. For a
diatomic ideal gas C p  72 R. Thus
Q
7
7
nRT   4.00 mol  8.31J/mol  K  60.0 K   6.98  103 J.
2
2
(b) The change in the internal energy is given by Eint = nCV T, where CV is the
specific heat at constant volume. For a diatomic ideal gas CV  52 R , so
Eint 
5
5
nRT   4.00 mol  8.31J/mol.K  60.0 K   4.99  103 J.
2
2
(c) According to the first law of thermodynamics, Eint = Q – W, so
W  Q  Eint  6.98  103 J  4.99  103 J = 1.99  103 J.
(d) The change in the total translational kinetic energy is
K 
3
3
nRT   4.00 mol  8.31J/mol  K  60.0 K   2.99  103 J.
2
2

19-78. (a) We use pV
i i  p f Vf


to compute :

ln V V  ln 1.0 10
ln  pi p f 
f

  5.
L 3
ln 1.0atm 1.0 105 atm
3
i
L 1.0 10
6
Therefore the gas is monatomic.
(b) Using the gas law in ratio form (see Sample Problem 19-1), the final temperature
is
T f  Ti
p f Vf
piVi
1.0 10
  273K 
5

atm 1.0 103 L
1.0atm  1.0 10
(c) The number of moles of gas present is
6
L

  2.7 10
4
K.



1.01105 Pa 1.0 103 cm3
piVi
n

 4.5 104 mol.
RTi
8.31 J/mol  K  273K 
(d) The total translational energy per mole before the compression is
Ki 
3
3
RTi   8.31 J/mol  K  273K   3.4 103 J.
2
2
(e) After the compression,
Kf 


3
3
RT f   8.31 J/mol  K  2.7 104 K  3.4 105 J.
2
2
2
 T , we have
(f) Since vrms
2
vrms,
i
2
rms,f
v

Ti
273K

 0.010.
T f 2.7 104 K