An ideal monatomic gas is at `P_(0), V_(0)`. It is taken to final volume `2V_(0)` when pressure is `P_(0)//2` in a process which is straight line on `P -V` diagram. Then

Options

A. The final temperature is greater than initial temperature
B. internal energy increases
C. the work doen by the gas is `+(P_(0)V_(0))/4`
D. The heat is absorbed in the process

Answer

D

Explanation

`(1) T_(A)=(P_(0)V_(0))/(nR)` and `T_(B)=(P_(0)//2.2 V_(0))/(nR)=(P_(0)V_(0))/(nR)`
`:. T_(A)=T_(B)`
(2) As `T_(A)=T_(B') DeltaU=0`
(3) workdone =area under the line AB with V
`=(P_(0)+P_(0)//2)/2 xx(2V_(0)-V_(0))=(3P_(0)V_(0))/4`
(4) As `DeltaU=0` and W gt 0 `rArr DeltaQ gt 0`