gas equations
DESCRIPTION
An introduction to ideal gas equations and examples of how to use them to solve problems.TRANSCRIPT
Gases
Gas equations
BOYLE’S LAWP
VP ∝ 1
have
PressureAmount
Gases
Volume
Temperature
Amount
x =x =x =x =x =
Volume(dm3)
Pressure(units)
k(proportionality constant: PV)
5 1 5
2,5 2 5
1,67 3 5
1,25 4 5
1 5 5
0,833 6 5
1
2
3
4
5
6 x =
Volume(dm3)
Pressure(units)
k(proportionality constant: PV)
V1 = 5 P1 = 1 5
2,5 2 5
1,67 3 5
1,25 4 5
1 5 5
0,833 6 5
1
2
3
4
5
6
V1 x P1 = k
Volume(dm3)
Pressure(units)
k(proportionality constant: PV)
V1 = 5 P1 = 1 5
V2 = 2,5 P2 = 2 5
1,67 3 5
1,25 4 5
1 5 5
0,833 6 5
1
2
3
4
5
6
V1 x P1 = k
V2 x P2 = k
V1 x P1 = k
V2 x P2 = k
V1 x P1 = V2 x P2
V1 x P1 = k
V2 x P2 = k
V1 x P1 = V2 x P2
P1V1 = P2V2
V1 x P1 = k
V2 x P2 = k
V1 x P1 = V2 x P2
P1V1 = P2V2
P1V1 = P2V2
PRESSURE AND TEMPERATUREP T
TP ∝
have
Amount
Gases
Temperature
Pressure
Volume
Amount
TemperaturePressure
(kPa) k =
(°C) (K)
20 293 1,00 0,0034
40 313 1,07 0,0034
60 333 1,14 0,0034
80 353 1,21 0,0034
100 373 1,28 0,0034
313 586 2,01 0,0034
1
2
3
4
5
6
TemperaturePressure
(kPa) k =
(°C) (K)
20 T1=293 P1=1,00 0,0034
40 313 1,07 0,0034
60 333 1,14 0,0034
80 353 1,21 0,0034
100 373 1,28 0,0034
313 586 2,01 0,0034
1
2
3
4
5
6
= k
TemperaturePressure
(kPa) k =
(°C) (K)
20 T1=293 P1=1,00 0,0034
40 T2=313 P2=1,07 0,0034
60 333 1,14 0,0034
80 353 1,21 0,0034
100 373 1,28 0,0034
313 586 2,01 0,0034
1
2
3
4
5
6
= k = k
= k = k=
=
CHARLES’ LAWV T
V ∝ T
TemperatureVolume
have
Gases
Amount
Pressure
TemperatureVolume
(cm3) k =
(°C) (K)
1 274 34 0,12
14 287 36 0,13
53 326 37 0,11
76 349 41 0,12
1
2
3
4
= k = k
= k = k=
=
GENERAL GAS EQUATION T
Amount
have
Gases
Pressure
TemperatureVolume
Pressure
TemperatureVolume
have
Gases
Amount
P1V1 = P2V2
==
=
QUESTION 1
Question
A gas bubble at the bottom of a lake has a volume of 200cm3 at a temperature of 8C. This bubble rises and when it reaches the surface where the temperature is 15C and the pressure is 100kPa, its volume increases to 550cm3. Calculate the pressure on the bubble when it was at the bottom of the lake.
8C = 281 K
15C = 288 K
==
P1 =
P1 = 268 kPa
==
P1 =
P1 = 268 kPa
==
P1 =
P1 = 268 kPa
have
Gases
Pressure
TemperatureVolume
Amount
Amount
Pressure
TemperatureVolume
have
Gases
Amount
(n)
moles(mol)
measured in
Amount
(n)
n
n
n
n
n
n
n
n
P
P
P
P
nP ∝
=
=
QUESTION
Question
1 mole of gas at 101,3 kPa and 273 K has a volume of 22,4 dm3. What is the volume of 5 moles of gas at 120 kPa and 280 K?
=
=
V2 =
V2 = 97 dm3
=
=
V2 =
V2 = 97 dm3
=
=
V2 =
V2 = 97 dm3
=
=
V2 =
V2 = 97 dm3
Standard Temperature: 273 K
Standard Pressure: 101,3 kPa
STP
Standard Temperature: 273 K
Standard Pressure: 101,3 kPa
STP
Standard Temperature: 273 K
Standard Pressure: 101,3 kPa
Volume of 1 mole of any gas at STP: 22,4 dm3
=
=
8,3 =
=
8,3 =
=
8,3 =
=
8,3 =
R =
PV = nRT
=
8,3 =
R =
PV = nRT
GENERAL GAS EQUATIONPV = nRT
Universal Gas Constant
R = 8,3 J • mol-1 • K-1
Universal Gas Equation
PV = nRT
=
=
8,3 =
R =
=
=
8,3 =
R =
𝑅=8,3 𝑘𝑃𝑎 ∙𝑑𝑚3
𝑚𝑜𝑙 ∙𝐾
1 000 Pa = 1 kPa
1 m3 = 1 000 dm3
1 000 Pa = 1 kPa1 m3 = 1 000 dm3
Universal Gas Constant
R = 8,3 J • mol-1 • K-1
Universal Gas Constant
R = 8,3 J • mol-1 • K-1
QUESTION
Question
What is the volume of 5 moles of gas at 120 kPa and 280 K?
PV = nRT
120•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120•V = 5•8,3•280
V =
V = 97 dm3
SI units
P : PaT : K
V : m3
n : mol
Question
What is the volume of 5 moles of gas at 120 kPa and 280 K?
120 kPa = 120 000 Pa
PV = nRT
120•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120 000•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120 000•V = 5•8,3•280
V =
V = 97 dm3
PV = nRT
120 000•V = 5•8,3•280
V =
V = 0,097 m3
PV = nRT
120 000•V = 5•8,3•280
V =
V = 0,097 m3
V = 0,097 m3 x = 97 dm3
QUESTION
Question
A sample of nitrogen gas has a mass of 7 g and a volume of 5,6 dm3 at a temperature of 27 °C. Calculate the pressure of the gas.
PV = nRT
N2 (g)
N2
M = 2(14)g•mol-1
M = 28 g•mol-1
n = 7 g N2 x = 0,25 mol N2
n = 7 g N2 x = 0,25 mol N2
T = 27 + 273 = 300 K
V = 5,6 dm3 x = 5,6 x 10-3 m3
V = 5,6 dm3 x = 5,6 x 10-3 m3
V = 5,6 dm3 x = 5,6 x 10-3 m3
T = 27 + 273 = 300 K
n = 7 g N2 x = 0,25 mol N2
V = 5,6 dm3 x = 5,6 x 10-3 m3
T = 27 + 273 = 300 K
n = 7 g N2 x = 0,25 mol N2
PV = nRT
P• 5,6 x 10-3 = 0,25 • 8,3 • 300
P =
P = 111 160 Pa
P = 111,160 kPa = 111 kPa
PV = nRT
P• 5,6 x 10-3 = 0,25 • 8,3 • 300
P =
P = 111 160 Pa
P = 111,160 kPa = 111 kPa
PV = nRT
P• 5,6 x 10-3 = 0,25 • 8,3 • 300
P =
P = 111 160 Pa
PV = nRT
P• 5,6 x 10-3 = 0,25 • 8,3 • 300
P =
P = 111 160 Pa
P = 111,160 kPa = 111 kPa