gas equations

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)

movie

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