If `y=e^(kt)` then `dy/dt` will be (A) `e^(kt) (B) e^(kt)/ k (C) te^(kt) (D) ke^(kt)`

If y=Ae^(-kt)cos(pt+c), then prove that (d^(2)y)/(dt^(2))+2k(dy)/(dx)+n^(2)y=0, where n^(2)=p^(2)+k^(2)

Show that , the solution x = e^(-kt) (a cos nt + b sin nt ), for all a and b , always satisfies the differenital equation (d^(2)x)/(dt^(2)) + 2k (dx)/(dt) + (k^(2) + n^(2)) x = 0

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. How many total degrees of freedom are present in H_(2) molecules in all types of motions ?

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. How many total degrees of freedom are present in H_(2) molecules in all types of motions ?

Consider the two statements Statement-l: y=sin kt satisfies the differential equation y'' + 9y = 0 . Statement 2 : y= e^(kt) satisfy the differential equation y'' + y' - 6y = 0 . The value of k for which both the statements are correct is

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. The rotational kinetic energy of H20 molecule is equal to

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. The vibrational kinetic energy of CO_2 molecule is

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. The rotational kinetic energy of H20 molecule is equal to