✤ ASSUMPTIONS

        1) Particles (molecules/atoms) of a gas always remain in motion in all possible direction.

        2) Collision of particles is elastic i.e. energy in collision remain constant.

        3) Particles of a Gas do not apply a force attraction or repulsion.

        4) Size of particle is zero as compare to distance b/w them.

✤ PRESSURE EXERTED BY GASONTHE WALL OF CONTAINER

                Let a Container is filled with Gas of mass

                    M & Volume=V

                Let no. a molecule per unit Volume = n

                Let mass of each atom be m. Let any instant

                    A(Vx, Vy, Vz)

                 For horizontal volume.

                            momentum=mVx

                 Let particle crebound with velocity-Ve

                            :. momentum = -mvx.

                 Change in momentum of atom = -2mvx

                 momentum-transferred to the wall = 2mVx

                                 Volume AXVxt

                 Total volume of atoms molecules) = nAVxt

                 Total momentum transferred to the wall = mVxnAVxt

                                                                               = mnAVx2t

                                                                               = mAVx2t

                 We know Force, F = MAVx2

                 We know P = F/A

                             P = MVx2

                 because Gas particle possessed any velocity along x, y, z axis

                   ✩ We should use ang. Velocity

                             P=MVx2

                    Here ² Vx²+Vy²+22

                    ATQ, that theory of KE

                            Vx²=Vy² =Vz2= avg velocity

                              V2 = 3Vx2    

                          Vx²=1/3V2

                :. Pressure = 1/3 MV2

                         NOTE:- Here V2 is Called r.m.s (c²)

                                       = 1/3 Mc²

                    Pressure per unit Volume:

                               P = 1/3 M/Vc2

                                = 1/3 Pc2

           We know, P=1/3 M/V c2

                 PV = MC2/3

            By Gas eq. PV = nRT

                  RT = Mc2/3

                 => C α √T

                 => C2 α T

✤ Avg. KE aasociated with gas :-

                We know, KE = 1/2 MC²

            KE per unit Volume = 1/2 Mc2/V

                      = 1/3 ρC²

✤ RELATION b/w K.E. & Pressure per unit Volume

          We Know

                P = 1/3 ρC² -①

                &     E = 1/2 ρC² -②

                      ①/②

                 P/E = 2/3

                  P= 2/3 E

                 P= 2/3 E or E = 3/2 P

✤ BOYLE'S LAW:- It stated that at constant temp.

                   PV = Constant;

              We knows P = 1/3 Mc2/V

                   PV= mc2/3

                If temp is constant than e is also constant ( C2 α T)

                      => MC²/3 = Constant

                       ie PV=Constant

                    which is Boyle’s law.

✤ CHARLE's LAW:- It states that at Constant pressure, & temperature is directly proportional to Volume

                We know

                      P = 1/3 Mc²/V

                      PV = mc²/3

                If P=constant

                              Vαc²

                         VαT i.e Charter's law.

✤ GAY-LUSSAC'S LAW :- It states that at constant Volume, the pressure exerted by a given mass of a gas is directly proportional to its absolute                                         temperature, i.e.,

                               P α T

                   We Know -

                             P= 1/3 M V’²/v

                              :. P α V²

                            V² α T :. P α T

✤ AVOGADRO'S LAW:- It states that equal volumes of all Gases under similar Conditions of temperatures & pressure Contain equal no. of molecules.

               We Know-

                     P = 1/3 mV²/v = 1/3 mnV²/v

                              P1 = P2

                    :. 1/3 m1n1V² = 1/3 m2 n2V²

                               m1 n1V²= m2 n2V² -①

               We know –

                         1/2 m1V1² = 1/3 m2V2²

                          m1V1²= m2V2² -②

                           Dividing ①/②

                    we get : n₁ = n₂

                   :. No. of molecules in Gas A = No. of molecules in Gas B

✤ GRAHAM'S LAW OF DIFFUSION: - It states that rate of diffusion of a gas is inversely proportional to square root of density.

   Pressure exerted by gas A ⇒ P₁ = ⅓ρ₁V₁²

   Pressure exerted by gas B ⇒ P₂ = ⅓ρ₂V₂²

  When steady state of diffusion is reached

      P₁ = P₂

    :. ⅓ρ₁V₁² = ⅓ρ₂V₂²

     (V₁/V₂)² = ρ₂/ρ₁

   V₁/V₂ = √ρ₂/ρ₁ (rate of diffusion ∝ r.m.s velocity)

      r₁/r₂ = √ρ₂/ρ₁

 → Pressure exerted by a Gas , P = ⅓V M vrms², = ⅓ PVrms²

✤ Degree of Freedom:- It represents the number of Independent quantities on Coordinates required the position of a dynamic System.

                       By the Exp. degree of freedom ---

                                f=3N-R Here, N = no. of particles

                                R = Relation b/w gives particles

              * For monoatomic Gas: * For di atomic Gas:-

                                    f=3 f=5

              * For triatomic Gas: For linear For non linear:-

                                    f=7 f=6

                          Energy associated with Gas:-

                                   ① translational

                                   ② Rotational

                                   ③ Vibrational.

✤ LAW OF EQUIPARTITION OF ENERGY:-

                 →   It state that many dynamical system in thermal equilibrium the energy is equally distributed amongst its various degrees of freedom

                       & the energy associate with each degree of freedom Per molecule is ½ KBT, where KB is Boltzmann's constant & T is absolute temp.

                       Average Kinetic energy of per molecule per f= ½ KBT.

                       Where KB= Boltzmann's Constant, T= absolute temp.

                 → Each square term in the total energy expression of a molecule contributes towards one degree of freedom.

                 → A monoatomic gas molecule has only translational kinetic energy 2 = 1/2 m v² + 1/2 m v² + 1/2 mv².

                       So a monoatomic gas molecule has only (translational) degrees of freedom.

✤ AVERAGE, ROOT MEAN SQUARE & MOST PROBABLE SPEED :-

                  Average Speed: It is defined as the arithmetic mean cay the speeds of the molecules of a gas at a given temperature.

                                      V= V₁+V₂+V₃....+V / n

                  By using Marvell distribution law,

                                 V = √8kBT/m = √8RT/m = √8PV/m

                  Root mean square speed: It is defined as the square root of the mean of the squares of the speeds of the individual molecules by Gas

                                 Vrms = √(V₁²+V₂²+V₃².....V / n

                 From Maxwell distribution law,

                             Vrms= √3kBT/m = √3RT/m = √3PV/m

                 Most Probable Speed: It is defined as the speed pressed by the max. no. of molecules in a gas sample at a given temp

                             Vmp= √2Kbt / M = √2RT / M = √2PV / M

                Relations b/w V, Vrms & Vmp

                            Vrms = √3kT/m = 1.73 √kT/m

                            V = √2kT/m = 1.60 √kT/m

        Vmp = √2 kT/m = 1.41 √kT/m

    Ratio Vrms: V: Vmp = 1.73 : 1.60 : 1.4

    Clearly, Vrms > V > Vmp

✤ For Your Knowledge:

      > The laws of equipartition of energy holds good for all degrees of freedom

      > A monoatomic gas molecule has only translational kinetic energy ,

                   KE = ½ mvx² + ½ mvy² + ½ mvz²

      See a monoatomic gas molecule has only three translational degrees of freedom.

      > In addition to KE, a diatomic molecule has two rotational kinetic energies.

                    Et+Er = ½ mvx² + ½ mvy² + ½ Iyωy² + ½ Izωz²

           Here the line joining the two atoms has been taken as X-axis about which there is no rotation . So the degree of freedom of a diatomic molecule is            5, it does not vibrate.

      > Diatomic molecule CO has a mode of vibration even at moderate temp. Its atom vibrate along the interatomic axis & contribute the vibrational energy          form Ev to the total energy.

                    E= Ex+Er+Ev

                        = ½mvx² + ½mvy² + ½mvz² + ½Ixyw² + ½Iyz² + ½kn² + ½kn²

                                   where k is the force constant of the oscillator, η the

                                   Vibrational Coordinate & η= d/dt

                               So, a diatomic molecule has 7 degree of freedom if it vibrates.

      > Each translational & rotational degree of freedom corresponds to one mode of absorption energy & has energy ½kBT. Each vibrational frequency has          two modes of energy (kinetic & potential) with corresponding energy equal in 3x ½kBT - kBT

 ✤ SPECIFIC HEAT

          1) for monoatomic: per degree of freedom    = ½ kBT

                                     Here degree of freedom = 3

                                               U= 3/2 kBT

                                               U= 3/2 RT

                                     we know, du= 3/2 R

               dT

      Cv = 3/2 R

      Cp-Cv=R

      Cp= R+Cv

           = R+ 3/2 R

      Cp= 5/2 R

            2) For diatomic : degree of freedom = 5

                                                  U= 5/2 RT

      Cv= dU/dT

      Cv= 5/2 R

     We know, Cp-Cv=R

      Cp = Cv+R

            = 5/2 R+R

      Cp = 7/2 R

            3) For triatomic : Gas degree of freedom in linear = 7

                                              ∴U = 7/2 RT

                                              Cv= dU/dT

      Cv= 7/2 R

       &

      Cp = 7/2 R+R

      Cp = 9/2 R

            4) For Polyatomic Gas:-

                                     Cv = f/2 R, Cp = (f/2+1) R

                                          γ = 1+2/f