Gas Laws - 1

The Gas Laws started .... in 1643 A.C. with invention of the barometer and continued until 1873 with the Vancer Waal's Equation. The gas law led to numerous concepts include the mole, temperatur, formula weight, absolute kinetic energy & stoichiometric coefficients. Gas laws were put forward by mainly three process. John Dalton, Robert Boyle, Jacques Charles. Modern kinetic molecules theory started with Bennoulli's suggestion in 1734 that pressure exerted by a gas on walls of its containes is sum of many collisions by individual molecules. THese gas laws have very significant role in practical life. Vander Waal'a Equation predicts critical temperature, pressure for liquefication of gases.

Measurement of ....:
Instrument used for measurement of pressure of gases c/d manometer. They are of two type:

(1) One in which longer is closed which is used for measuring pressure less than atmospheric pressure.

(2) One in which longer limb is open which is used for measuring pressure greater than atmospheric pressure.

1 atm = 76 cm = 760 mm or 760 ...

1 atm = 1.0132 bar      1 bar = 0.987 atm      1 Pa = 1 Nm-2      1 bar = 105 Pa

Pressure: It is force per unit area.
Pressure of menometer = gh

Dumb Question: How P = gh in manometer ?
Ans: Suppose height of mercury column = h cm
Area of cross-section of tube = cm²
Vol. of mercury column = A x h cm³
Then mass of mercury column. 6 = A x h x gm
Weight of mercury column = (A x h x ) x g = Force
Pressure (P) =

Illustration: Manometer is connected to a gas containing bulb. Open arm reads 43.7 cm whereas arm connected to bulb reads 15.6 cm. If barometer pressure is 743 mm mercury. What is pressure of gas in bulb.
Ans: Note: heights are measured always from bottom. Diff. of mercury levels in two arms = 43.7 - 15.6 = 28.1 cm. As level in limb connected to bulb is lower than that of open limb, this means pressure of gas is more than atm. pressure.
Pressure of gas = Atm. pressure + Diff. in levels of mercury
                         = 74.3 cm + 28.1 cm
                         = 102.4 cm
                         =
                         = 1.36 s bar

Measurement of Temperature:

0C = 5/9(F0 - 32)

K = 0C + 273.1 s

C Celcius
F Farhenheit
K Kelvin

Scals read same value ?
Ans: Suppose both read same value as 'x'
⁰C = 5/9(F - 32) Br /> x = 5/9(x - 32) => x = - 40⁰C

II. Gas Laws:

(a) Boyle's Law: At constant temperatue, volume of a given mass of gas is inversely proportional to its pressure.
      V

=>  PV = constant

Let P₁ & V₁ are initial pressure & volume of gas & P₂ & V₂ are final pressure & volume at constant temlerature.
P₁V₁ = P₂V₂    at constant temperature.
In other words.
at constant temperature gas density is directly proportional to pressure.

Dumb Question: How density is proportional to pressure ?
Ans:   d =
And PV = K,
      d =
=> d P

Question: gas at one atm. pressure is connected to evcuated bulb of 0.5 L capacity through stop-clock. On opening, pressure constant temperature. What is volume of bulb X ?
Ans: Suppose volume of bulb 'X' = VL
i.e. V₁ = V & P₁ = 1 atm
After connecting to bulb of 0.5 L
Volume = (V + 0.5) L
i.e. V₂ = (V + 0.5) L,    P₂ = 570 mm = 570/760 atm.
By P₁V₁ = P₂V₂
1 X V = x (v + 0.5)
On solving V = 1.5 L

(b) Charle's ⁰ Law: It states that pressure remaining constant, volume of given mass of gas increase or decrease by 1/293 of its volume at 0⁰C for every 1⁰C rise or fall in temperature.

V_t = V₀ + x t      t ⁰C
at constant pressure, volume of given mass of gas is directly proportional to its temperature in degrees kelvin.
V T
= constant    at constant pressure is directly proportional to its temperature in degrees kelvin.
at constant volume.
at constant volume.


Illustration: An open vessel contain 200 mg of air at 17⁰C. What % of wt of air would be expelled if vessel is heated to 117⁰C.
Ans: Suppose volume of 200 mg of air at 17⁰C = V ml
As pressure remains constant.

Dumb Question: Why pressure is constant ?
Ans: Because vessel is open & only atm. pressure is acting.
V₂ = 1.34 V
Volume of air expelled = 1.34 V - V = 0.34 V
   Mass of 1.34 V air at 117⁰C = 200mg
   Mass of 0.34 V air at 117⁰C = x 0.34 mg
Mass % of air expelled = x 100 = 25.37 %