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Gas Pressure in the light of
KMT
• Pressure is defined as the force the gas exerts
on a given area of the container in which it is
contained.
𝑃 =
𝐹
𝐴
2
Units of Pressure
• SI unit of pressure is Pascal
P =
𝑁
𝑚2
= 𝑃𝑎
• A Pascal is defined as the force of 1 Newton (N) spread
over an area of 1 m2
• OTHER UNTIS:
• 1 atm = 101325 Pa = 101325 Nm-2
• 1 atm = 101.325 KPa
• 1 atm = 14.7 Psi (Pounds per square inch)
• 1 atm = 760 torr = 760 mm of Hg
• 1 J = 1 Nm = 107ergs = 1 Kgm2s-2
• 1 Cal = 4.18 J
• 1 atm = 1.01325 bar
3
Effect of change in Pressure on the
Volume of a Gas – Boyle’s Law
Pressure and volume are
inversely related at constant
temperature.
𝑃 ∝
1
𝑉
𝑃 =
𝐾
𝑉
𝑃𝑉 = 𝐾
𝑃1 𝑉1 = 𝐾
𝑃2 𝑉2 = 𝐾
𝑃1 𝑉1 = 𝑃2 𝑉2 4
“Father of Modern Chemistry”
Robert Boyle
Boyle’s Law: P1V1 = P2V2
Boyle’s Law – Isotherms
• When P of a gas is plotted againstV at different
temperature, hyperbola curves are obtained which are
called Isotherms.
• As the temperature increases, the isotherms goes away
from both the axis.This is due to increase in volume at
higher temperature.
6
Effect of change in Temperature on
the Volume of gas – Charles’s Law
• At constant Pressure, volume of given
mass of a gas increase or decreases by
decreases by 1/273 times of its original
original volume at 0oC for every 1oC rise
7
Jacques-Alexandre Charles
Charles’ Law: V1/T1 = V2/T2
Derivation of Critical Form of
Charles’s Law
• Suppose the volume of a gas at 0oC =Vo
• Volume at 1oC = 𝑉𝑜 = 𝑉𝑜 + 𝑉𝑜
1
273
• Volume at 2oC = 𝑉𝑜 = 𝑉𝑜 + 𝑉𝑜
2
273
• Volume at toC = 𝑉𝑡 = 𝑉𝑜 + 𝑉𝑜
𝑡
273
𝑉𝑡 = 𝑉𝑜 1 +
𝑡
273
𝑉𝑡 = 𝑉𝑜
273+𝑡
273
𝑡 + 273 = 𝑇 𝐾𝑒𝑙𝑣𝑖𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
𝑉𝑡 = 𝑉𝑜
𝑇
273
𝑉𝑡 =
𝑉𝑜
273
𝑇 9
Charles’s law - Isobar
When volume of gas is
plotted against
temperature at different
pressures, a straight line is
obtained.
Each constant pressure line
is called Isobar.
10
Absolute Zero
• According to Charles’s law: At constant Pressure,
volume of given mass of a gas increase or decreases
decreases by 1/273 times of its original volume at 0oC
11
Absolute Zero
• According to Charles’s law: At constant Pressure,
volume of given mass of a gas increase or decreases
decreases by 1/273 times of its original volume at 0oC
12
Absolute Zero
• At exact -273 oC the volume of a
given mass of gas reduces to zero.
• -273oC = 0K
• The temperature at which the
given volume of a gas reduces to
Zero is called Absolute Zero i.e. 0K
or -273oC.
• Actually all the gases liquefy or
solidify before they reach -273oC.
• This temperature is considered to
as the lowest possible
temperature
13
Avogadro’s Law
• At constant temperature
and pressure, the volume
of a gas is directly related
to the number of moles.
𝑉 ∝ 𝑛
𝑉 = 𝐾𝑛
𝑉1
𝑛1
=
𝑉2
𝑛2
14
Amedeo Avogadro
15
Avogadro’s Law: V1/n1=V2/n2
Avogadro’s Law – Molar Volume
• It has been calculated that if we have 1 dm3 of H2
gas, its mass at STP will be 0.09 g
• ∴ 0.09 g H2 at STP = 1 dm3
• 2.016 g of H2 at STP =
1
0.09
× 2.106 = 22.414 𝑑𝑚3
• 2.016 g H2 = 1 mol
• Hence 1 mol H2 at STP will occupy volume 22.414
dm3
• This volume is called Molar volume.
16
Avogadro’s Law
• Equal volumes of all gases at same
temperature and pressure must contains
equal number of molecules
17
Ideal Gas Equation
• An equation that shows the effect of simultaneous
changes in pressure and temperature on the
volume of a given gas is called Ideal Gas equation.
• Ideal gas equation is the combination of three gas
laws:
• Boyle’s Law
• Charles’s Law
• Avogadro’s Law
18
Derivation of Ideal Gas Equation
• According to Boyle’s law: 𝑉 ∝
1
𝑃
… . . (𝑖)
• According to Charles’s law: 𝑉 ∝ 𝑇 … . . (𝑖𝑖)
• According to Avogadro’s law: 𝑉 ∝ 𝑛 … . . (𝑖𝑖𝑖)
• Combining equations (i) (ii) and (iii) we get:
𝑉 ∝
𝑛𝑇
𝑃
𝑉 =
𝑛𝑅𝑇
𝑃
𝑃𝑉 = 𝑛𝑅𝑇
Where R is called General Gas constant.
• If n=1, then PV = RT or
𝑃𝑉
𝑇
= 𝑅
• So,
𝑃1 𝑉1
𝑇1
= 𝑅
• And
𝑃2 𝑉2
𝑇2
= 𝑅
• ∴
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
19
Significance of Ideal Gas Equation
• Calculation of Molecular mass of gas:
𝑃𝑉 = 𝑛𝑅𝑇
𝑛 =
𝑊
𝑀
Where W = mass of gas and M= Molecular mass
𝑃𝑉 =
𝑊
𝑀
𝑅𝑇
𝑀𝑃𝑉 = 𝑊𝑅𝑇
𝑀 =
𝑊𝑅𝑇
𝑃𝑉
• Calculation of Density of gas:
𝑃𝑉 =
𝑊
𝑀
𝑅𝑇
𝑀𝑃 =
𝑊
𝑉
𝑅𝑇
As
𝑊
𝑉
= 𝑑
𝑀𝑃 = 𝑑𝑅𝑇
𝑑 =
𝑀𝑃
𝑅𝑇
20
sidra.javedali@gmail.com
21

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Gas Laws

  • 1. 1
  • 2. Gas Pressure in the light of KMT • Pressure is defined as the force the gas exerts on a given area of the container in which it is contained. 𝑃 = 𝐹 𝐴 2
  • 3. Units of Pressure • SI unit of pressure is Pascal P = 𝑁 𝑚2 = 𝑃𝑎 • A Pascal is defined as the force of 1 Newton (N) spread over an area of 1 m2 • OTHER UNTIS: • 1 atm = 101325 Pa = 101325 Nm-2 • 1 atm = 101.325 KPa • 1 atm = 14.7 Psi (Pounds per square inch) • 1 atm = 760 torr = 760 mm of Hg • 1 J = 1 Nm = 107ergs = 1 Kgm2s-2 • 1 Cal = 4.18 J • 1 atm = 1.01325 bar 3
  • 4. Effect of change in Pressure on the Volume of a Gas – Boyle’s Law Pressure and volume are inversely related at constant temperature. 𝑃 ∝ 1 𝑉 𝑃 = 𝐾 𝑉 𝑃𝑉 = 𝐾 𝑃1 𝑉1 = 𝐾 𝑃2 𝑉2 = 𝐾 𝑃1 𝑉1 = 𝑃2 𝑉2 4 “Father of Modern Chemistry” Robert Boyle
  • 6. Boyle’s Law – Isotherms • When P of a gas is plotted againstV at different temperature, hyperbola curves are obtained which are called Isotherms. • As the temperature increases, the isotherms goes away from both the axis.This is due to increase in volume at higher temperature. 6
  • 7. Effect of change in Temperature on the Volume of gas – Charles’s Law • At constant Pressure, volume of given mass of a gas increase or decreases by decreases by 1/273 times of its original original volume at 0oC for every 1oC rise 7 Jacques-Alexandre Charles
  • 9. Derivation of Critical Form of Charles’s Law • Suppose the volume of a gas at 0oC =Vo • Volume at 1oC = 𝑉𝑜 = 𝑉𝑜 + 𝑉𝑜 1 273 • Volume at 2oC = 𝑉𝑜 = 𝑉𝑜 + 𝑉𝑜 2 273 • Volume at toC = 𝑉𝑡 = 𝑉𝑜 + 𝑉𝑜 𝑡 273 𝑉𝑡 = 𝑉𝑜 1 + 𝑡 273 𝑉𝑡 = 𝑉𝑜 273+𝑡 273 𝑡 + 273 = 𝑇 𝐾𝑒𝑙𝑣𝑖𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑉𝑡 = 𝑉𝑜 𝑇 273 𝑉𝑡 = 𝑉𝑜 273 𝑇 9
  • 10. Charles’s law - Isobar When volume of gas is plotted against temperature at different pressures, a straight line is obtained. Each constant pressure line is called Isobar. 10
  • 11. Absolute Zero • According to Charles’s law: At constant Pressure, volume of given mass of a gas increase or decreases decreases by 1/273 times of its original volume at 0oC 11
  • 12. Absolute Zero • According to Charles’s law: At constant Pressure, volume of given mass of a gas increase or decreases decreases by 1/273 times of its original volume at 0oC 12
  • 13. Absolute Zero • At exact -273 oC the volume of a given mass of gas reduces to zero. • -273oC = 0K • The temperature at which the given volume of a gas reduces to Zero is called Absolute Zero i.e. 0K or -273oC. • Actually all the gases liquefy or solidify before they reach -273oC. • This temperature is considered to as the lowest possible temperature 13
  • 14. Avogadro’s Law • At constant temperature and pressure, the volume of a gas is directly related to the number of moles. 𝑉 ∝ 𝑛 𝑉 = 𝐾𝑛 𝑉1 𝑛1 = 𝑉2 𝑛2 14 Amedeo Avogadro
  • 16. Avogadro’s Law – Molar Volume • It has been calculated that if we have 1 dm3 of H2 gas, its mass at STP will be 0.09 g • ∴ 0.09 g H2 at STP = 1 dm3 • 2.016 g of H2 at STP = 1 0.09 × 2.106 = 22.414 𝑑𝑚3 • 2.016 g H2 = 1 mol • Hence 1 mol H2 at STP will occupy volume 22.414 dm3 • This volume is called Molar volume. 16
  • 17. Avogadro’s Law • Equal volumes of all gases at same temperature and pressure must contains equal number of molecules 17
  • 18. Ideal Gas Equation • An equation that shows the effect of simultaneous changes in pressure and temperature on the volume of a given gas is called Ideal Gas equation. • Ideal gas equation is the combination of three gas laws: • Boyle’s Law • Charles’s Law • Avogadro’s Law 18
  • 19. Derivation of Ideal Gas Equation • According to Boyle’s law: 𝑉 ∝ 1 𝑃 … . . (𝑖) • According to Charles’s law: 𝑉 ∝ 𝑇 … . . (𝑖𝑖) • According to Avogadro’s law: 𝑉 ∝ 𝑛 … . . (𝑖𝑖𝑖) • Combining equations (i) (ii) and (iii) we get: 𝑉 ∝ 𝑛𝑇 𝑃 𝑉 = 𝑛𝑅𝑇 𝑃 𝑃𝑉 = 𝑛𝑅𝑇 Where R is called General Gas constant. • If n=1, then PV = RT or 𝑃𝑉 𝑇 = 𝑅 • So, 𝑃1 𝑉1 𝑇1 = 𝑅 • And 𝑃2 𝑉2 𝑇2 = 𝑅 • ∴ 𝑃1 𝑉1 𝑇1 = 𝑃2 𝑉2 𝑇2 19
  • 20. Significance of Ideal Gas Equation • Calculation of Molecular mass of gas: 𝑃𝑉 = 𝑛𝑅𝑇 𝑛 = 𝑊 𝑀 Where W = mass of gas and M= Molecular mass 𝑃𝑉 = 𝑊 𝑀 𝑅𝑇 𝑀𝑃𝑉 = 𝑊𝑅𝑇 𝑀 = 𝑊𝑅𝑇 𝑃𝑉 • Calculation of Density of gas: 𝑃𝑉 = 𝑊 𝑀 𝑅𝑇 𝑀𝑃 = 𝑊 𝑉 𝑅𝑇 As 𝑊 𝑉 = 𝑑 𝑀𝑃 = 𝑑𝑅𝑇 𝑑 = 𝑀𝑃 𝑅𝑇 20