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What is the Basic 3 law of an ideal gas ? – An Overview


Perfect gas :- A perfect gas or an ideal gas may be define as a stable of substance whose evaporation from its liquid state is complete known as perfect or ideal gas.

Important topics

  1. Boyle’s Law
  2. Charle’s Law
  3. Gay Lussac Law
  4. General Gas Equation
  5. Avagadro’s Law
  6. Universal Gas Constant
  7. Regnault’s Law
  8. Relationshhip between specific heat and Cp and Cv
  9. Problems
  10. Triple Points Of Water
  11. Thermal Expansion
  12. Question And Answer

Boyle’s law

This law was founded by Rovest Boyle in 1662. This law state the absolute pressure at given mass of perfect gas varies in inversely as its volume provide the constant temperature.

                   P  ∝ 1/V

                   PV = constant

The more useful form of the above equation

  P1V1 = P2V2 = P3V3 ……………… = constant.

Charle’s law

This law was founded by French mam AC charle’s in 1787. This law may be state in the following two ways :-

A)The volume is given mass of a perfect gas varies directly proportional to absolute temperature provided the pressure remain constant

                    V ∝ T

                   V/ T = constant

        V1/T1 = V2/T2 = V3/T3 ………… = constant.

All perfect gas change in volume by1/273th of its original volume of 0oC far every 1oC change in temperature , when the pressure remain constant.

B) At a given volume the pressure of a given mass of a perfect gas is directly proportional to its absolute temperature.

                             Vt = V0  + 1/273 V0 * t

                             Vt = V0 [273+t/273]

                             Vt = V0/T0

Gay lussac law

The absolute pressure of a perfect gas is directly proportional to its absolute temperature when its volume remains constant.

                   P   ∝  T

                   P/T = constant

          P1/T1 = P2/T2 = P3/T3 ………… constant.

General gas equation

It is derived by consider the fact that all the free variable i.e, pressure , volume , and temperature change simutanously.

General gas equation  (Ideal gas equation)

                   PV = nRT

Boyle’s law = V  ∝ 1/P

Charle’s law = V  ∝ T

Avogadro hypotheses :- V  ∝ n

                             V  ∝  nT/P

                             VP  ∝ nT

PV = nRT

Avagadro’s law

This law state equal volume of all gasses will have equal number of molecules under identical condition of temperature and pressure remain constant.

                 Volume of gass . number of molecules

                         V  ∝  n           (constant temperature and pressure)

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 Universal gas constant

 The universal gas constant or molar constant  of a gasis the product of the gas constant and the molecular  of gas

                   Ru = MR

Where m = molecular mass of gas in kg mole.

                  R = gas constant

                   Ru = M1R1 = M1R1 …..

Regnault’s law

This law state the specific heat Cp (specific heat at constant pressure) and Cv (specific heat at constant volume) of a gas do not change with the change of temperature and pressure.

Specific heat :– The amount of heat required to raise the temperature of its unit mass through 1o

Specific heat at constant pressure :- It is the amount of heat required to raise the temperature of a unit mass of a gas through 1o , when its pressure is kept constant.

                             Qp = mCp ( T2 – T1)

Specific heat at constant volume :

It is the amount of heat required to raised the temperature of a unit mass of  a gas through 1o , when its volume kept constant.

                   Qv = mCv ( T2 – T1)

                   SI unit = 0.72kj/kg

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Relationshhip between specific heat and Cp and Cv

Let us consider in kg of a gas enclosed in a container and is being heated at constant pressure.


T1 = initial temperature of the gas

T2 = final temperature of the gas

V1 = intial volume of the gas

V2 = final volume of the gas

Cp = specific heat at constant pressure

Cv = specific heat at constant volume

P  = absolute constant pressure

 We know that heat supplied to the gas at constant pressure.

                   Q = mCp ( T2 – T1)  ………………………………… 1

We also know that a part of heat is utilized in doing external work , while the remaining part of heat is used in increasing the internal energy of gas.

                   Q = w + du ……………………………………………… 2

Now heat utilized for external work.

                   W = P (V2 – V1)  ………………………………………..  3

And increasing in internal energy .

                   Dv = mCv (T2 – T1) ……………………………………..  4

Putting the value

of w equation dv in equation 2

Q = P (V2 – V1)  + mCv (T2 – T1) ………………………………….  5

 Using characterstic gas equation.

We have ,

For intial condition

                   Pv1 = MRT1

                   Pv2 = MRT2   …………… for final compare

      P (V2 – V1)  = MR (T2 – T1) ……………………………………  6

Putting the equation of P (V2 – V1)  from eqution 6 in equation 5

Now we get ,

                   Q = MR (T2 – T1) + mCv (T2 – T1)

Now putting the value of Q in equation 1

                   mCp (T2 – T1) = MR (T2 – T1) + mCv (T2 – T1)

in absence equation we get ,

                   Cp = R + Cv

                   Cp – Cv = R

The equation may be written as

                   Cv (Cp/Cv – 1) = R

                   Cv = R/v – 1

The value of R is taken as 287 J/kg or 0.287 wJ/kg

 The equation may be also written

                   Cp = Cv + R

Dividing both side by Cv we get

                   Cv/Cp = 1 + R/Cv


 3m3 of a gas at 7 bar abs pressure expands to a volume of 12m3 at constant temperature. Find the final pressure of the gas

Ans.Given :- V1 = 3m3  ,  P1 = 7 bar  ,  V2 = 12m3

Temperature = constant

Let P2 is the final pressure of the gas

We know that

                  P1V1 = P2V2

         P2  =  p1V1/V2    =  7*3/12

                            = 1.75 bar abs.

 Determine the volume occupied by a given mass of air at a temperature of 245oC , if the same mass of air occupies 3m3 at 32oC . the pressure of the air remains unchanged.

Ans.Given :- T1 = 23oC = 296 k  ,  T2  = 245oC = 578 k

V1 = 2m3  ,  pressure = constant

Let V2 is the volume occupied by air at 245oC

We know that ,

                  (charle’s law)

                  V2 = V1T2/T1 = 2*518/296

                            V2 = 3.5m3

 A certain gas occupies 2m3 at a temperature of 127oC. the pressure of the gas is 7 bar and the gas expand in such as manner that the volume becomes 8m3 and the temperature 15oC. what will be the pressure of gas after expansion ?

 Ans.Given :- V1 = 2m3 , T1 = 127o C = 400 k

                  P1 = 7 bar , V2 = 8m3 , T2 = 15o C = 288 k

Let P2 is the pressure of gas after expansion.

Now we know that ,

                            P1V1/T1 = P2V2/V2 (gas equation)

                            P2 = P1V1T2/V2T1

= 7*2*288/8*400

P2 = 1.26 bar.

Triple point of water

The triple point of water is the state at which the three phase of water namely ice , liquid , water vapour are equally stable and co-exist in equilibrium.  It is unique because it occurs at a specific temperature of 273.16k and a specific pressure of 0.46 cm of Hg column. Thus for water ,

                           Ptr = 273.16k or 0.01oc

Thermal expansion

Almost all solids , liquids and gases expand on heating. The increase in the size of a body when it is heated is called thermal expansion.

Different typres of thermal expansion :-

  1. Linear expansion :- It is the increase in the length of a metal rod on heating.
  2. Superficial expansion :- It is the increase in the surface area of a metal sheet on heating.
  3. Cubical expansion :- It is the increase in the volume of block on heating.


Q. Why do solids expand on heating ?

Ans. Cause of thermal expansion  :- All solids consist of atoms and molecules. At any given temperature , these atoms and molecules are held at equilibrium  distance by force of attraction.
When a solid is heated , the amplitude of vibration of its atoms and molecules increases. The average inter-atomic separation  increases. The results in the thermal expansion of the solid.

Q2. What is latent heat ? give its units.

Ans. Latent heat :- When a solid change into liquid or a liquid changes into gas , it absorbs heat. But this heat does not show up an increase in temperature. This heat , used to change the state , is hidden  or latent and is therefore called latent heat.

Q3. Define term water equivalent. Give its CGS and SI units.

Ans. Water equivalent :- The water equivalent of a body is define as the mass of water which requires the same amount of heat as is required by the given body for the same rise of temperature.
        Water equivalent = mass * specific heat
Or             w = mc
                 The CGS unit of water equivalent is g and the SI unit is Kg.

Q4. What is melting and boiling point ?

Ans. Melting point :- The temperature at which the solid and the liquid states of a substance co-exist in thermal equilibrium with each other is called its melting point.
Boiling point :– the temperature at which the liquid and vapour states of a substance co-exist in thermal equilibrium with each other is called its boiling point.

Q5. what does the ideal gas law state ?

Ans. Ideal gas law :- The ideal gas law states that the pressure, temperature, and volume of gas are related to each other.

Q6. Who made the ideal gas law ?

Ans. Development of the ideal gas law :-The pressure, volume, temperature, and amount of an ideal gas are related by one equation that was derived through the experimental work of several individuals, especially Robert Boyle, Jacques A.C. Charles, and Joseph Gay Lussac.

Q7. what is the ideal gas law used for?

Ans. The ideal gas law relates the four independent physical properties of a gas at any time. The ideal gas can be used in Stoichiometry problems in which chemical reaction involve gases.



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