Saved Bookmarks
| 1. |
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» In Isothermal process , K_i = d p / d v_1 / vK_i = PIN adiabatic CONDITION , K_a = d p / d v_2 / vK_a = γ P .Now : K_a / K_i = d v_1 / d v_2 = γSince the VALUE of γ is greater than 1 .i.e. γ > 1 .= > Therefore , the value of d v_1 > d v_2 . |
|