JEE Main20205 Sep 2020Morning ShiftPhysicsThermodynamicsActual
Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as A → B , A → C and A → D . The change in internal energies during these process are taken as E A B , E A C and E A D and the work done as W A B , W A C and W A D . The correct relation between these parameters are:
Options
- AE A B = E A C < E A D , W A B > 0 , W A C = 0 , W A D < 0
- BE A B = E A C = E A D , W A B > 0 , W A C = 0 , W A D < 0
- CE A B < E A C < E A D , W A B > 0 , W A C > W A D
- DE A B > E A C > E A D , W A B < W A C < W A D
Correct answer
B. E A B = E A C = E A D , W A B > 0 , W A C = 0 , W A D < 0
Step-by-step solution
E A B = E A C = E A D d U = n f R 2   T f - T i W A B > 0 ( + )   as   V ↑ W A C = 0   as   V = constant W A D < 0 ( - )   as   V ↓ Δ T is same for E A B = E A C = E A D