III. Ideal gases
Equation of state
PV  nRT
PV 
R  8.31 J /( mol  K )
m

RT
n = m/μ
n - number of moles
m - total mass of gas
μ=M - molar (atomic) mass (“weight”)
PV
 const
T
P1V1 P2V2

T1
T2
1. Isotherms (Boyle’s law): T=const
P
PV=const
T2 > T1
P1V1= P2V2
T1
V
2. Isobars (Charles’s law): P=const
V
V/T=const
V1/T1= V2/T2
P1
P2 > P1
V
T(K)
-273°C
T(C)
3. Isochors (Gay-Lussaec ): V=const
P
V1
P/T=const
V2 > V1
T(K)
P1/T1= P2/T2
Example 1: The temperature of an amount of an ideal gas has been
increased twice, while the volume has been increased four times.
What happened with the pressure?
T2 = 2T1
V2 = 4V1
P2V2 P1V1

T2
T1
P2 V1 T2 1
1

 2 
P1 V2 T1 4
2
P2 /P1 - ?
P1
P2 
2
Example 2: What is the volume of 1 mole of an ideal gas at
“standard temperature and pressure”?
n = 1 mol
T = 273 K (0° C)
P = 1 atm = 1.013x105 Pa
V-?
PV  nRT
V  nRT / P
R = 8.31 J/(mol *K)
V=22.4x10-3 m3 = 22.4 L
IV. Kinetic theory
1. Avogadro number and number of moles
One mole is the amount of substance that contains as many elementary entities
as there are atoms in 12g of carbon -12
N A  6.022  10 23 molecules / mol
N
n
NA
2. Molecular weight
mtot  mN  mnN A  n
mN A  
n - number of moles
N - number of molecules
NA - Avogadro number
mtot - total mass
m - mass of one molecule
μ=M - molar (atomic) mass (“weight”)
3. Boltzmann constant
R
kB 
NA
nR  NkB
kB  1.38  1023 J / molecules  K
4. Kinetic energy and temperature of an Ideal gas)
Definition:
There are no interactions between atoms (molecules) of an ideal gas
Macro: PV  nRT  Nk B T
2
2
2
m

v

Micro: PV  K tr  N
3
3
2
m  v2  3
3
K tr  N
 nRT  Nk B T
2
2
2
 
 v 2  v 2
K tr ~ T
ave
v rms
!
3k B T
3RT
 v  

m

2
Example: The atomic weights of nitrogen (N2) is 28 g/mol, and the atomic weight
of oxygen (O2) is 32 g/mol. Let us conceder these gases at normal conditions.
a) What is the average translational kinetic energy of a single N2 molecule?
b) What is the average translational kinetic energy of one mol of N2?
c) What is the ratio of the average translational kinetic energy of N2 and O2?
d) What is the rms velocity of N2?
e) What is the ratio of the rms velocities of N2 and O2?
  N 2   28 g / mol
 O2   32 g / mol
a ) K tr1  N 2   ?
b) K tr N A   N 2   ?
c) K tr  N 2  / K tr O2   ?
d )v rms  N 2   ?
e)v rms  N 2  / v rms O2   ?
K tr1 
3
3
k B T  1.38  10  23  J / K   273K  565  10  23 J
2
2
K tr N A  
3
N A k B T  N A K tr1  3400 J
2
K tr N 2  / K tr O2   1
vrms N 2  
3RT


vrms N 2  / vrms O2  
3  8.31 J /( mol  K )  273K
3
28 10 kg / mol
 O2 
32
8


 N 2 
28
7
 500m / s
5. Kinetic theory of Ideal gas*
N/2
vx t  x
N/2
2
F
N
N vx x N
2
mv
2
PV  V 
fx  m
 m2vx2  N
 Ktr
A
2
2
t
2
3
2
3
N
F
f
2
V  A x
v
f m x
t
vx  2vx
x
 vx
t

vx2  v y2  vz2  13 vx2  v y2  vz2
vx2  13 v 2
2 mv2 2
PV  N
 K tr
3
2
3

Question
The value of vrms for monatomic He
gas at 300 K is 1370 m/s (!). In order
to double vrms, the temperature must
be changed to ___ K.
1. 150
2. 425
3. 600
4. 1200