A diatomic ideal gas is compressed adaibatically 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

D

Explanation

For an adiabatic process, the temperature-volume
relationship is
`T_1V_1^(gamma-1)=T_2V_2^(gamma-1) rArr T_1=T_2((V_2)/(V_1))^(gamma-1)`
Here `gamma=1.4 (for diatomic gas). V_2=(V_1)/32, T_1=T_i, T_2=aT_i`
`:. T_i=aT_i[1/32]^(1.4-1) =aT_i[1/(2^5)]^0.4=(aT_i)/4 :. a=4`