When `1mol` of a monoatomic ideal gas at `TK` undergoes adiabatic change under a constant external pressure of `1atm`, changes volume from `1 L to 2L`. The final temperature (in K) would be

Options

A. `T/(2^((2//3)))`
B. `T+2/3xx0.0821`
C. T
D. `T-2/3xx0.0821`

Answer

A

Explanation

`TV^(gamma-1)=`Constant ( `because` change is adiabatic)
`T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)`
For monoatomic gas `gamma` = `5/3`
`thereforeT_(1)V_(1)^(2//3)=T_(2)V_(2)^(2//3)rArrT(1)^(2//3)=T_(2)(2)^(2//3)`
`T_(2)=T/(2^((2//3)))`