Question

The first law of thermodynamics was given as `q=DeltaU+(-w)`, where `q` is heat given to a system and `DeltaU` represents increase in internal energy and `-w` is work done by the system. Various processes such as isothermal, adiabatic, cyclic, isobaric and isochoric process in terms of `I` law of thermodynamics leads for important results. The molar heat capacity for `1` mole of monoatomic gas is `3/2 R` at constant volume and `5/2 R` at constant pressure.
A system is allowed to move from state `A` to `B` following path `ACB` by absorbing `80J` of heat energy. The work done by the system is `30J`. The work done by the system in reaching state `B` from `A` is `10J` through path `ADB,`
Which statements are correct?
(1) Increase in internal energy from state `A` to state `B` is `50J`. (2) If path `ADB` is followed to reach state `B, DeltaU=50J`. (3) If work done by the system in path `AB` is `20J,` the heat absorbed during path `AB=70J`. (4) The value `U_C - U_A` is equal to `U_D - U_B`. (5) Heat absorbed by the system to reach `B` from `A` through path `ADB` is `60J`.
A. `1,5`
B. `1,3,5`
C. `1,2,3,5`
D. `1,4,5`

Answer 1

Correct Answer - c
`ACB=AC+BC`
Heat absorbed `80J`
Work done by the system `=30 J`
`:. W=-30`
`:. U_(B)-U_(A)=50 J`
`q=PDeltaV`
`ADB=AD+BD`
Work done by the system `=10 J`
`:. W=-10J`
In path `AB,U_(B)-U_(A)=50 J`
Also `w=-10 J`
`:. q=50+10=60 J`