THERMODYNAMICS

Answer :

Each molecule has f degree of freedom

                             C_p = [(f / 2) + 1]R,          C_v = (f/2)R

                             (C_p / C_v) = [({(f/2) + 1}R) / {(f/2)R}]

                                         = [1 + {1 / (f/2)}] = [1 + (2/f)]

Answer :

Area of ellipse = π ∙ a ∙ b.

                   The diagram can be treated as ellipse.

                   here a = [(P₂ – P₁) / 2]    &       b = [(V₂ – V₁) / 2]

                   ∴ area = π ∙ a ∙ b

                             = (π/4)(P₂ – P₁)(V₂ – V₁)

                   magnitude of work done = area covered in a cyclic process

                                                          = (π/4)(P₂ – P₁)(V₂ – V₁)   

Answer :

given (C_p / C_v) = γ

                             C_p – C_v = R

                   ∴       [(C_p / C_v) – 1] = [R / (CV)]

                   ∴       C_v = [R / (γ – 1)]………(1)

                   As p ∙ Δv = n ∙ R ∙ ΔT

                   ∴ ΔT = [(P ∙ ΔV) / (n ∙ R)]

                   given : volume changes from V to 2V

                   ∴       ΔV = 2V – V = V

                   ∴       ΔT = (PV / nR)……….(2)

                   change in internal energy = ΔU = n ∙ C_v ∙ ΔT

                   from (1) & (2)

                             ΔU = n × [R / (γ – 1)] × (PV / nR)

                             ΔU = [PV / (γ – 1)]   

Answer :

T₁ = 927 + 273 = 1200 K

                   T₂ = 27 + 273 = 300 K

                   internal energy ∝ Temperature

                             U ∝ T

                             (U₁ / U₂) = (T₁ / T₂)

                   ∴       [(U₁ – U₂) / U₂] = [(T₁ – T₂) / T₂]

                   ∴       % change in energy = [{(1200 – 300) / (300)} × (100)]

                                                           = 3 × 100 = 300% 

Answer :

For all gases C_p – C_v = R

                   given C_p – C_v = a for hydrogen

                             C_p – C_v = b for oxygen

                             hence    a = b = R