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

Options

A. 3
B. 4
C. 5
D. 6

Answer

B

Explanation

For an adiabatic change, `TV^(gamma-1)` = constant
and for a diatomic gas, `gamma=7/5`
`T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)andT_(2)=alphaT_(1)`
`thereforeT_(2)/T_(1)=(V_(1)/V_(2))^(gamma-1)=(V/(V//32))^(7/5-1)`
`therefore(alphaT_(1))/T_(1)=(32)^(2//5)=(2^(5))^(2//5)=4" "thereforealpha=4`