An ideal gas of mass `m` in a state `A` goes to another state `B` via three different processes as shown in Fig. If `Q_(1), Q_(2)` and `Q_(3)` denote the heat absorbed by the gas along the three paths, then

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 ot the gas first increases and then decreases continously in path (1).

Answer

A, B, C

Explanation

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