Question

The volume of an ideal gas is V at a pressure P. On increasing the pressure by dP, the change in volume of the gas is (dV1) under isothermal conditions and (dV2) under adiabatic conditions. Is dV1 >dV2 or vice-versa and why?​

Answer 1

Answer:

In Isothermal process ,

K_i = d p / d v_1 / v

K_i = P

In adiabatic condition ,

K_a =  d p / d v_2 / v

K_a = γ P .

Now :

K_a / K_i = d v_1 / d v_2 = γ

Since the value of γ is greater than 1 .

i.e. γ > 1 .

= > Therefore , the value of d v_1 > d v_2 .