P - Gainesville ISD

Report
Ideal Gas Law
Ideal Gas Equation
This equation is a combination of 3 simpler gas relationships:
Boyle’s law: V a 1 (at constant n and T)
P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
nT
P
nT
nT
V = constant x
=R
P
P
R is the gas constant
PV = nRT
What
does
R=?
The conditions 0 0C and 1 atm are called
standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.
PV = nRT
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
L atm
R = 0.082057 mol K
Ideal Gas Problem #1
What is the volume (in liters) occupied by 49.8 g of HCl at a
temperature of 25.0°C and a pressure of 0.950 atm?
T = 25.0 + 273.15 = 298.15 K
PV = nRT
nRT
V=
P
P = 0.950 atm
n = 49.8 g x
1.37 mol x 0.0821
V=
V = 35.3 L
L•atm
mol•K
1 mol HCl
= 1.37 mol
36.45 g HCl
x 298.15 K
0.950 atm
At STP this same mass of HCl
occupied a volume of 30.6 L
Ideal Gas Problem #2
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = 273.15 K
PV = nRT
nRT
V=
P
P = 1 atm
n = 49.8 g x
1.37 mol x 0.0821
V=
V = 30.6 L
L•atm
mol•K
1 mol HCl
= 1.37 mol
36.45 g HCl
x 273.15 K
1 atm
In the last question, this same mass of HCl at
25°C and 0.95 atm had a volume of 35.3 L
Ideal Gas Problem #3
What is the pressure (in atmospheres) of a gas in a 6.00 L
container filled with 25.0 g of carbon dioxide at 23.5°C?
T = 23.5°C+273.15 = 296.65 K
PV = nRT
nRT
P=
V
P=
V = 6.00 L
1 mol CO2
n = 25.0 g x
= 0.568 mol
44.01 g CO2
L•atm
0.568 mol x 0.0821 mol•K
x 296.65 K
P = 2.31 atm
6.00 L
Ideal Gas Problem #4
What is the temperature (in °C) of a gas in a 2.00 L container filled
with 15.0 g of nitrogen with a pressure of 6.75 atm?
P = 6.75 atm
PV = nRT
PV
T=
nR
V = 2.00 L
1 mol N2
n = 15.0 g x
= 0.535 mol
28.01 g N2
6.75 atm x 2.00 L
T=
L•atm
0.535 mol x 0.0821 mol•K
T = 307.35 K – 273.15 = 34.2°C
Ideal Gas Problem #5
What mass of oxygen gas occupies a volume of 1.50 L at 15.0 °C
and 3.25 atm of pressure?
T = 15.0°C+273.15 = 288.15 K
PV = nRT
V = 1.50 L
P = 3.25 atm
PV
n=
RT
n=
3.25 atm x 1.50 L
0.0821 L•atm x 288.15 K
mol•K
n = 0.200 mol
32.00 g O2 = 6.41 g O
0.200 mol
2
1 mol O2
Ideal Gas Problem #6
(Making additional conversions)
What mass of oxygen gas occupies a volume of 765 mL
at 12.3 °C and 333 kPa of pressure?
T = 12.3°C+273.15 = 285.45 K
PV = nRT
V = 765 mL
P = 333 kPa
PV
n=
RT
n=

1 L 

 1000 mL 


 1 atm 


 101.3 kPa 


= 0.765 L
= 3.29 atm
3.29 atm x 0.765 L
0.0821 L•atm x 285.45 K
mol•K
n = 0.107 mol
32.00 g O2 = 3.44 g O
0.107 mol
2
1 mol O2
Ideal Gas Problem #7
(Making additional conversions)
What mass of radioactive radon gas occupies a volume of 439 mL
at 101.9 °C and 950 mmHg of pressure?
T = 101.9°C+273.15 = 375.05 K





1L

-4 L
V = 439 mL
=
4.39x10

1106 m L 
PV = nRT
P = 950 mmHg
PV
n=
RT
n=
1.25 atm x 4.39x10-4 L
0.0821 L•atm x 375.05 K

1 atm 

 760 mmHg 


= 1.25 atm
n = 1.782x10-5 mol
mol•K
1.782x10-5
222 g Rn
mol
1 mol Rn
= 4.00x10-3 g Rn
Before & After Calculations
In some situations, we know the amounts of all 4 variables and
our task is to determine one of them under new conditions where
one or more of the others are changing.
Here is how we use
PV=nRT for this situation:
= constant
Since PV
nT
We can use it to represent a “before” and “after”
set of conditions like this:
P1 V1
P V
= 2 2
n1 T1
n2 T2
1 = before or initial and 2 = after or final
Ideal Gas Problem #8
(Before & After)
Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb
containing argon at 1.20 atm and 18°C is heated to
85°C at constant volume. What is the final pressure
of argon in the lightbulb (in atm)?
P1 V1
P2 V2
=
n1 T1
n2 T2
Therefore:
n, and V are constant so they
can be left out of the equation.
P1
P2
=
T1
T2
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
Dalton’s Law
of Partial
Pressures
Mixtures of gases require special
handling when doing calculations.
Collected gas
Evaporated
water
Ptotal = P + P
Collected gas
Evaporated
water
Bottle being filled with oxygen gas
NaClO3
Bottle full of oxygen
gas and water vapor
2 NaClO3 (s)  2 NaCl (s)
PT = PO + PH
2
+ 3 O2 (g)
2
O
Dalton’s Law of Partial Pressures
V and T
are
constant
P1
P2
Ptotal = P1 + P2
Pressure of Collected
Gas Alone
Pcollected gas = Ptotal - PH2O
Total
pressure of
gas in
container
Vapor
pressure of
water at this
temperature
Consider a case in which two gases, A and B, are in a
container of volume V.
nART
PA =
V
nA is the number of moles of A
nBRT
PB =
V
nB is the number of moles of B
PT = PA + PB
c
c
PB =
PA =
A PT
Pi =
cP
i
T
nA
A=
nA + nB
c
c
nB
B =
nA + nB
B PT
mole fraction (
c ) = nn
i
i
T
Gas Problem #9
A sample of natural gas contains 8.24 moles of CH4, 0.421 moles
of C2H6, and 0.116 moles of C3H8. If the total pressure of the
gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi =
c
cP
i
propane
T
=
PT = 1.37 atm
0.116 C H
= 0.0132
+ 0.421C H + 0.116C H
3
8.24CH
4
2
8
6
3
8
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
Chemistry in Action:
Scuba Diving and the Gas Laws
P
Depth (ft)
Pressure
(atm)
0
1
33
2
66
3
V

similar documents