Certain perfect gas is found to obey the law `PV^(3//2)`= constant, during adiabatic process. If such a gas at initial temprerature `T` is adiabatically compressed to half of the initial volume, its final temperature will be

Options

A. `2Tsqrt2`
B. 4T
C. `Tsqrt2`
D. 2T

Answer

C

Explanation

`PV^(3/2)`=constant, Again `P=(nRT)/V`
`therefore(nRT)/VxxV^(3/2)`=constant (K)
or, `TV^(1/2)=K/(nR)=K.`(constant)
For two states, `T_(1)V_(1)^(1/2)=T_(2)V_(2)^(1/2)`
`T_(2)=T_(1)((V_(1))/(V_(2)))^(1//2)` , But `V_(2)=(V_(1))/2` (given)
`thereforeT_(2)=T_(1)((V_(1))/(V_(1/2)))^(1/2)=T_(1)sqrt2`