10/3/2013
IDEAL GAS LAW
P V = n R T
Brings together gas properties.
C be
Can
b derived
d i d from
f
experiment
i
andd
theory.
P = pressure in atm V = volume in L
n = number of moles T = temp. in K
R = gas constant = 0.0821 L atm
mol K
Using PV = nRT
How much N2 is required to fill a small room with a
volume of 960 cubic feet to P = 745 mm Hg at 25 oC?
R = 0.0821 L•atm/K•mol
Solution
1. Get all data into pproper
p units
3
3
V = 960 ft 1 m
1003 cm3 1 L = 27,180 L
3
3
3.281 ft 1 m3
1000 cm3
o
T = 25 C + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg)
= 0.980 atm
Using PV = nRT
R = 0.0821 L•atm/
L•atm/K•mol
Solution
1. V = 27,180 L T =298 K P = 0.980 atm R = .0821
2. Now calculate
calculate-n = PV / RT
n = 1090 mol or 30,500 g of N2
1
10/3/2013
Gases and Stoichiometry
2 H2O2(l) --->
---> 2 H2O(g) + O2(g)
If you decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the pressure of O2 at
25 oC? Of H2O?
Gases and Stoichiometry
2 H2O2(l) --->
---> 2 H2O(g) + O2(g)
Solution
1.1 g H 2 O 2 •
1 mol
34.0 g
0.032 mol H 2O2 •
= 0.032 mol
1 mol O 2
= 0.016 mol O 2
2 mol H2 O 2
P of O 2 = nRT/V
=
(0.016 mol)(0.0821 L • atm/K • mol)(298 K)
2.50 L
P of O2 = 0.16 atm
What is P of H2O? Could calculate as above. But
recall Avogadro’s hypothesis.
V ∝ n at same T and P
P ∝ n at same T and V
There are 2 times as many moles of H2O as moles
of O2. P is proportional to n. Therefore, P of H2O
is twice that of O2.
P of H2O = 0.32 atm
2
10/3/2013
Molecular Formula and Ideal Gas Law
•
A compound contains only nitrogen and hydrogen and
is 84.7 % nitrogen by mass. A gaseous sample of the
compound has a density of 0.996 g/L at 710. torr and
100.0C.
What is the molecular formula of the compound?
Solving the problem:
1. Assume 1.00 L and therefore = .996 g
2. Find moles of each element to find empirical formula
84.7 g N 1 mol = 6.05 mol N
14.0 g
15.3 g H 1 mol H =15.1 mol H
1.01 g H
6.05/6.05 = 1
15.1/6.05 =2.5
Empirical Formula = N2H5 efm = 33.05 g/mol
:
Finishing the problem
2. Find moles of gas using ideal gas law…
P = .934 atm
V=1.00 L R = 0.0821 T = 373 K
n = PV = 0.0305 mol
RT
44.
5.
6.
7.
Find Molecular Weight = M.W.
M W = .996
996 g/ .0305
0305 mol
M.W. = 32.66 g/mol
Divide M.W./e.f.m. = 32.66/33.05 = 1
Therefore Molecular Formula = N2H5
Dalton’s Law
John Dalton
1766--1844
1766
*sometimes gases are bubbled
through water to collect
Pt = Pdry gas + Pwater vapor
Pwater vapor is temperature dependent
3
10/3/2013
Dalton’s Law of Partial Pressures
2 H2O2(l) --->
---> 2 H2O(g) + O2(g)
0.32 atm
0.16 atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = P(H
( 2O)) + P(O
( 2) = 0.48 atm
Dalton’s Law: total P is sum of PARTIAL pressures.
When pressure and temperature are constant,
Dalton’s Law states that the partial pressures are
equal to the mole ratios of the gases!
Dalton’s Law Expanded
•
The partial pressure of CH4(g) is 0.175 atm and
that of O2(g) is 0.250 atm in a mixture of the
two gases.
A. What is the mole fraction of each gas in the mixture?
Since the mol and pressure are directly related…
.175 mol CH4 = .412 CH4
.250 mol O2 = .588 O2
.425 mol
.425 mol
B. If the mixture occupies 10.5 L at 650C, calculate the
total number of moles of gas in the mixture.
P = .425 atm V = 10.5 L T = 338 K R = .0821
n = PV/RT n = .161 mol of gas total
Finishing the problem
C. Using the information from the last problem,
how many grams of each gas is present in the
mixture?
• n = .161 mol total
• mol CH4 = (.412) (.161mol) = .0663 mol CH4
• 0.0663 mol CH4 16.0 g = 1.06 g CH4
1 mol
• mol O2 = (.588) (.161mol) = .0947 mol O2
• 0.0947 mol O2 32.0 g = 3.03 g O2
1 mol
4
10/3/2013
GAS DENSITY
PV = nRT
n
P
=
V
RT
m
P
=
M• V
RT
where M = molar mass
d and M proportional
5