charles’s law + pressure law

7/31/2019 Charles’s Law + pressure law

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Charles’s Law

For a fixed mass of gas, the VOLUME isDIRECTLY PROPORTIONAL to

TEMPERATURE (when the PRESSURE isCONSTANT)

V α T , P is constant

V/T = constant (The unit of T is Kelvin)

2

2

1

1

T

V

T

V



CHARLES’ LAW

Why does the air trapped in capillary tube

has been heated by immerse it in the water?



Charles’ Law

T (oC)

V (cm-3)

0 100

10

40

30

20

-300 -200 -100

-273

O K = - 273oC >>>> Absolute Zero of Temperature



T (oK)

V (cm-3)

0 100 200 300

10

40

30

20

O K = - 273oC >>>> Absolute Zero of Temperature

Kelvin Scale of Temperature



Pressure Law

For a fixed mass of gas, the PRESSUREis DIRECTLY PROPORTIONAL to

TEMPERATURE (when the VOLUME isCONSTANT)

P α T , V is constant

P/T = constant (The unit of T is Kelvin)

P1

= P2

T1 T2



PRESSURE LAW



Pressure Law

T /oC

P (x 10 5)

/Pa

0 100

2

1

-300 -200 -100

-273

O K = - 273oC > Absolute Zero of Temperature



T (oK)

P (x105)

/Pa

0 100

2

1

O K = - 273oC > Absolute Zero of Temperature

Kelvin Scale of Temperature



• The air in a foot pump has an initial volume of

2800 cm3 and pressure 100 kPa. The outlet of

the pump is closed and the piston pushed

inwards until the volume of the air becomes 700

cm3. what is the pressure of the compressed air

in the pump?

solutionP1 = 100 kPa

P2 = ???

V1 = 2800 cm3

V2 = 700 cm3

Using Boyle’s Law

P1V1 = P2V2

100 x 2800 = P2 x 700

P2 = 400 kPa



• An iron cylinder containing gas has a pressure of 360 kPa

when it is kept in a store at temperature 27oC. What is the

pressure of the gas when cylinder is moved outdoors where

the temperature is 40o

C?

2

2

1

1

T

P

T

P

solution P1 = 360 kPa

P2 = ???

T1 = (27 + 273) = 300 KT2 = (40 + 273) = 313 K

Using Pressure Law

P2 = 375.6 kPa

313300

3602P



A syringe in a refrigerator contains 4.5 cm3 of air at

-3oC. When the syringe was taken out and placed

in a room where the temperature was 27oC, the air

in it expands. Calculate the final volume of the air

in the syringe.

2

2

1

1

T

V

T

V

solutionV1 = 4.5 cm3

V2 = ???T1 = (-3 + 273) = 270 KT2 = (27 + 273) = 300 K

Using Charles’ Law

V2 = 5.0 cm3

300270

5.42P



Phenomenon of Gas Law

1.Bau busuk macamammonia le..

2.Bahaya tak?

3.Kalau terhidu je, takle membunuh

4.Tapi bau air ni yangmacam tercemar.

Panggil Karamsingh

Walia nak? Mana taukat hulu sungai ni adababi berkubang

5.Ye le..bau najismughalazah al kubra,kita nak samak badankat mana lepas ni?

6. Kuala lumpur la…

PV = constant

Q: What is the volume of bubble at Q if the volume of bubble at P is 6 cm3.(P atm = 10 m water)

P

Q1 m

4 m




1.Kau selalu

kemekkan bolaselepas kalah…

Kalau menang kautak jugak beli bola

2.Rebus dalam airpanas je…

3.Selepas masak

kau nak makan ke?Selepas ni kalau kaukemekkan lagi, akukasi kau menang je..

4.Tak melawan takbest le…

Q: The air pressure in a ping pong ball at 27 oC is 120,000 Pa.The ball is heated in hot water at100 oC. What is the air pressure in the ball now?

P = constant

T




Di tepi pantaiyang indah…

ku baringseorang diri…

suasana…ngeri

Terlihat dikau diwaktu malam…

wajahmu selaluku terbayang..

Q: The volume of air trapped in a straw at 27 oC is 5 cm3. Calculate the temperature of thetrapped air if its volume increases to 6 cm3. State your answer in Kelvin and degree Celsius.

V = constant

T

charles’s law + pressure law

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