Inversion temperature `(T_(i)=(2a)/(Rb))` is defined as the temperature above which if gas is expanded
adiabatically it gets warm up but if temperature of gas is lower than `T_(i)` then it will cool down. What will
happen to gas if it is adiabatically expanded at `50^(@)C` if its Boyle's temperature is `20^(@)C`

Options

A. heating
B. cooling
C. constant
D. None of these

Answer

B

Explanation

Boyle temperature, `T_(b)= (a)/(Rb)= 20^(@) C= 293 K`
inversion temperature , `T_(i)= (2a)/(Rb)= 586K = 313^(@)C`
`therefore" "` at `50^(@)Clt T_(i)`, on expansion cooling occurs.