Question

Two samples A and B of same gas have same Initial pressure and volume. A undergoes an Isothermal expansion and B is subjected to adiabatic expansion. If the final volume in both cases is double the initial volume and work done In both the cases are same, then prove that 2^1-γ - 1 = In2, γ = \(\frac{c_p}{c_v}\).

Answer 1

Let the initial pressure, p_i = p₁

initial volume, V_i = V₁

final volume, V_f = V₂

= 2V₁

Work done in isothermal process,

W_iso = 2.303 RT log \((\frac{v_2}{v_1})\)

Since PV = RT (1 mol of gas),

W_iso = P₁ V₁ ln₂ ......(1)

work done in adiabatic process,

For adiabatic process,

P₁ \({V_1^γ}\) =P1 \(V_2^γ\) K (constant)

Since work done is same for both processes, W_iso = W_ad

Re-arranging and simplifying (3),

2^1-γ - 1 = (1 - γ) ln2.