A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is

Answer

`a = 4`

Explanation

For adiabatic process,
`TV^(gamma-1) = cosntant , T_(2) = T_(1) ((V_(1))/(V_(2)))^(gamma-1)`
`T_(2) = T_(1)(32)^((7)/(5)-1) , T_(2) = 4T_(1) rArr a = 4`