A gas undergoes change in its state from position `A` to position `B` via three different path as shown in Fig. Select the correct alternatives :

(A)Change in internal energy in all the three paths is equal
(B)In all the three paths heat is absorbed by the gas
(C)Heat absorbed/released by the gas is maximum in path 1
(D)Temperature of the gas first increases and then decreases in path 1

Options

A. Change in internal energy in all the three paths is equal.
B. In all the three paths heat is absorbed by the gas .
C. Heat absorbed `//` released by the gas is maximum in path `(1)`
D. Temperature of the gas first increases and then decreases continuously in path `(1)`

Answer

A, B, C

Explanation

Internal energy `(U)` depends only on the initial and final states. Hence. `Delta U` will be same in all the three paths. In all the three paths, work done by the gas is positive and the product `PV` or temperature `T` is increasing . Therefore, internal energy is also increasing. So, from the first law of thermodynamics, heat will be absorbed b the gas. Further, area under `P-V` graph is maximum in path 1 while `DeltaU` is same for all the three paths . Thereforr, heat absorbed by the gas is maximum in path 1. For temperature of the gas, we can see that product `PV` first increases in path 1 but whether it is decreasing or increasing later on we cannot say anything about it unless the exact values are known to us.